How do I check if a string contains a specific word?

Asked 2023-09-20 20:26:52 View 306,848

Consider:

$a = 'How are you?';

if ($a contains 'are')
    echo 'true';

Suppose I have the code above, what is the correct way to write the statement if ($a contains 'are')?

Answers

Now with PHP 8 you can do this using str_contains:

if (str_contains('How are you', 'are')) { 
    echo 'true';
}

Please note: The str_contains function will always return true if the $needle (the substring to search for in your string) is empty.

$haystack = 'Hello';
$needle   = '';

if (str_contains($haystack, $needle)) {
    echo "This returned true!";
}

You should first make sure the $needle (your substring) is not empty.

$haystack = 'How are you?';
$needle   = '';

if ($needle !== '' && str_contains($haystack, $needle)) {
    echo "This returned true!";
} else {
    echo "This returned false!";
}

Output: This returned false!

It's also worth noting that the new str_contains function is case-sensitive.

$haystack = 'How are you?';
$needle   = 'how';

if ($needle !== '' && str_contains($haystack, $needle)) {
    echo "This returned true!";
} else {
    echo "This returned false!";
}

Output: This returned false!

RFC

Before PHP 8

You can use the strpos() function which is used to find the occurrence of one string inside another one:

$haystack = 'How are you?';
$needle   = 'are';

if (strpos($haystack, $needle) !== false) {
    echo 'true';
}

Note that the use of !== false is deliberate (neither != false nor === true will return the desired result); strpos() returns either the offset at which the needle string begins in the haystack string, or the boolean false if the needle isn't found. Since 0 is a valid offset and 0 is "falsey", we can't use simpler constructs like !strpos($a, 'are').

Answered   2023-09-20 20:26:52

  • @DTest - well yes of course it will return true because the string contains 'are'. If you are looking specifically for the word ARE then you would need to do more checks like, for example, check if there is a character or a space before the A and after the E. - anyone
  • Very good comments above! I never use != or ==, after all !== and === is best option (in my opinion) all aspect considered (speed, accuracy etc). - anyone
  • @jsherk Why not regexes, then? Something like " are ". - anyone
  • As for not catching 'care' and such things, it is better to check for (strpos(' ' . strtolower($a) . ' ', ' are ') !== false) - anyone
  • I tend to avoid this issue by always using strpos($a, 'are') > -1 to test for true. From a debugging perspective, I find my brain wastes fewer clock cycles determining if the line is written correctly when I don't have to count contiguous equals signs. - anyone

You could use regular expressions as it's better for word matching compared to strpos, as mentioned by other users. A strpos check for are will also return true for strings such as: fare, care, stare, etc. These unintended matches can simply be avoided in regular expression by using word boundaries.

A simple match for are could look something like this:

$a = 'How are you?';

if (preg_match('/\bare\b/', $a)) {
    echo 'true';
}

On the performance side, strpos is about three times faster. When I did one million compares at once, it took preg_match 1.5 seconds to finish and for strpos it took 0.5 seconds.

Edit: In order to search any part of the string, not just word by word, I would recommend using a regular expression like

$a = 'How are you?';
$search = 'are y';
if(preg_match("/{$search}/i", $a)) {
    echo 'true';
}

The i at the end of regular expression changes regular expression to be case-insensitive, if you do not want that, you can leave it out.

Now, this can be quite problematic in some cases as the $search string isn't sanitized in any way, I mean, it might not pass the check in some cases as if $search is a user input they can add some string that might behave like some different regular expression...

Also, here's a great tool for testing and seeing explanations of various regular expressions Regex101

To combine both sets of functionality into a single multi-purpose function (including with selectable case sensitivity), you could use something like this:

function FindString($needle,$haystack,$i,$word)
{   // $i should be "" or "i" for case insensitive
    if (strtoupper($word)=="W")
    {   // if $word is "W" then word search instead of string in string search.
        if (preg_match("/\b{$needle}\b/{$i}", $haystack)) 
        {
            return true;
        }
    }
    else
    {
        if(preg_match("/{$needle}/{$i}", $haystack)) 
        {
            return true;
        }
    }
    return false;
    // Put quotes around true and false above to return them as strings instead of as bools/ints.
}

One more thing to take in mind, is that \b will not work in different languages other than english.

The explanation for this and the solution is taken from here:

\b represents the beginning or end of a word (Word Boundary). This regex would match apple in an apple pie, but wouldn’t match apple in pineapple, applecarts or bakeapples.

How about “café”? How can we extract the word “café” in regex? Actually, \bcafé\b wouldn’t work. Why? Because “café” contains non-ASCII character: é. \b can’t be simply used with Unicode such as समुद्र, 감사, месяц and 😉 .

When you want to extract Unicode characters, you should directly define characters which represent word boundaries.

The answer: (?<=[\s,.:;"']|^)UNICODE_WORD(?=[\s,.:;"']|$)

So in order to use the answer in PHP, you can use this function:

function contains($str, array $arr) {
    // Works in Hebrew and any other unicode characters
    // Thanks https://medium.com/@shiba1014/regex-word-boundaries-with-unicode-207794f6e7ed
    // Thanks https://www.phpliveregex.com/
    if (preg_match('/(?<=[\s,.:;"\']|^)' . $word . '(?=[\s,.:;"\']|$)/', $str)) return true;
}

And if you want to search for array of words, you can use this:

function arrayContainsWord($str, array $arr)
{
    foreach ($arr as $word) {
        // Works in Hebrew and any other unicode characters
        // Thanks https://medium.com/@shiba1014/regex-word-boundaries-with-unicode-207794f6e7ed
        // Thanks https://www.phpliveregex.com/
        if (preg_match('/(?<=[\s,.:;"\']|^)' . $word . '(?=[\s,.:;"\']|$)/', $str)) return true;
    }
    return false;
}

As of PHP 8.0.0 you can now use str_contains

<?php
    if (str_contains('abc', '')) {
        echo "Checking the existence of the empty string will always"
        return true;
    }

Answered   2023-09-20 20:26:52

  • @Alexander.Plutov second of all you're giving me a -1 and not the question ? cmon it takes 2 seconds to google the answer google.com/… - anyone
  • +1 Its a horrible way to search for a simple string, but many visitors to SO are looking for any way to search for any of their own substrings, and it is helpful that the suggestion has been brought up. Even the OP might have oversimplified - let him know of his alternatives. - anyone
  • Technically, the question asks how to find words not a substring. This actually helped me as I can use this with regex word boundries. Alternatives are always useful. - anyone
  • +1 for the answer and -1 to the @plutov.by comment because , strpos is just a single check meanwhile regexp you can check many words in the same time ex: preg_match(/are|you|not/) - anyone
  • Regular Expressions should be the last resort method. Their use in trivial tasks should be discouraged. I insist on this from the height of many years of digging bad code. - anyone

Here is a little utility function that is useful in situations like this

// returns true if $needle is a substring of $haystack
function contains($needle, $haystack)
{
    return strpos($haystack, $needle) !== false;
}

Answered   2023-09-20 20:26:52

  • @RobinvanBaalen Actually, it can improves code readability. Also, downvotes are supposed to be for (very) bad answers, not for "neutral" ones. - anyone
  • @RobinvanBaalen functions are nearly by definition for readability (to communicate the idea of what you're doing). Compare which is more readable: if ($email->contains("@") && $email->endsWith(".com)) { ... or if (strpos($email, "@") !== false && substr($email, -strlen(".com")) == ".com") { ... - anyone
  • @RobinvanBaalen in the end rules are meant to be broken. Otherwise people wouldn't come up with newer inventive ways of doing things :) . Plus have to admit I have trouble wrapping the mind around stuff like on martinfowler.com. Guess the right thing to do is to try things out yourself and find out what approaches are the most convenient. - anyone
  • Another opinion: Having an utility function which you can easily wrap can help debugging. Also it loundens the cry for good optimizers which eliminate such overhead in production services. So all opinions have valid points. ;) - anyone
  • Of course this is usefull. You should encourage this. What happens if in PHP 100 there is a new and faster way to find string locations ? Do you want to change all your places where you call strpos ? Or do you want to change only the contains within the function ?? - anyone

To determine whether a string contains another string you can use the PHP function strpos().

int strpos ( string $haystack , mixed $needle [, int $offset = 0 ] )`
<?php

$haystack = 'how are you';
$needle = 'are';

if (strpos($haystack,$needle) !== false) {
    echo "$haystack contains $needle";
}

?>

CAUTION:

If the needle you are searching for is at the beginning of the haystack it will return position 0, if you do a == compare that will not work, you will need to do a ===

A == sign is a comparison and tests whether the variable / expression / constant to the left has the same value as the variable / expression / constant to the right.

A === sign is a comparison to see whether two variables / expresions / constants are equal AND have the same type - i.e. both are strings or both are integers.

One of the advantages of using this approach is that every PHP version supports this function, unlike str_contains().

Answered   2023-09-20 20:26:52

  • If i use "care" its return true as well :( - anyone

While most of these answers will tell you if a substring appears in your string, that's usually not what you want if you're looking for a particular word, and not a substring.

What's the difference? Substrings can appear within other words:

  • The "are" at the beginning of "area"
  • The "are" at the end of "hare"
  • The "are" in the middle of "fares"

One way to mitigate this would be to use a regular expression coupled with word boundaries (\b):

function containsWord($str, $word)
{
    return !!preg_match('#\\b' . preg_quote($word, '#') . '\\b#i', $str);
}

This method doesn't have the same false positives noted above, but it does have some edge cases of its own. Word boundaries match on non-word characters (\W), which are going to be anything that isn't a-z, A-Z, 0-9, or _. That means digits and underscores are going to be counted as word characters and scenarios like this will fail:

  • The "are" in "What _are_ you thinking?"
  • The "are" in "lol u dunno wut those are4?"

If you want anything more accurate than this, you'll have to start doing English language syntax parsing, and that's a pretty big can of worms (and assumes proper use of syntax, anyway, which isn't always a given).

Answered   2023-09-20 20:26:52

  • this should be the canonical answer. Because we're looking for words and not substrings, regex is appropriate. I'll also add that \b matches two things that \W doesn't, which makes it great for finding words in a string: It matches beginning of string (^) and end of string ($) - anyone
  • this should be the correct answer.. the rest of the answers will find "are" in a string like "do you care".. As mentioned by @Dtest - anyone
  • @RobertSinclair Is that so bad? If you asked me if the string "do you care" contains the word "are" I would say "yes". The word "are" is clearly a substring of that string. That's a separate question from """Is "are" one of the words in the string "do you care"""". - anyone
  • @Paulpro Eventhough OP didn't specify the $a is a phrase, I'm pretty sure it was implied. So his question was how to detect the Word inside the Phrase. Not if a Word contains a Word inside of it, which I would assume would be irrelevant more often than not. - anyone
  • @Jimbo it does works, you're just missing the `\` 3v4l.org/ZRpYi - anyone

Look at strpos():

<?php
$mystring = 'abc';
$findme   = 'a';
$pos = strpos($mystring, $findme);

// Note our use of ===. Simply, == would not work as expected
// because the position of 'a' was the 0th (first) character.
if ($pos === false) {
    echo "The string '$findme' was not found in the string '$mystring'.";
} else {
    echo "The string '$findme' was found in the string '$mystring',";
    echo " and exists at position $pos.";
}

Answered   2023-09-20 20:26:52

Using strstr() or stristr() if your search should be case insensitive would be another option.

Answered   2023-09-20 20:26:52

  • A note on the php.net/manual/en/function.strstr.php page: Note: If you only want to determine if a particular needle occurs within haystack, use the faster and less memory intensive function strpos() instead. - anyone
  • @tastro Are there any reputable benchmarks on this? - anyone
  • This might be slower, but IMHO strstr($a, 'are') is much more elegant than the ugly strpos($a, 'are') !== false. PHP really needs a str_contains() function. - anyone
  • It blows my mind that this is not the accepted answer - anyone

Peer to SamGoody and Lego Stormtroopr comments.

If you are looking for a PHP algorithm to rank search results based on proximity/relevance of multiple words here comes a quick and easy way of generating search results with PHP only:

Issues with the other boolean search methods such as strpos(), preg_match(), strstr() or stristr()

  1. can't search for multiple words
  2. results are unranked

PHP method based on Vector Space Model and tf-idf (term frequency–inverse document frequency):

It sounds difficult but is surprisingly easy.

If we want to search for multiple words in a string the core problem is how we assign a weight to each one of them?

If we could weight the terms in a string based on how representative they are of the string as a whole, we could order our results by the ones that best match the query.

This is the idea of the vector space model, not far from how SQL full-text search works:

function get_corpus_index($corpus = array(), $separator=' ') {

    $dictionary = array();

    $doc_count = array();

    foreach($corpus as $doc_id => $doc) {

        $terms = explode($separator, $doc);

        $doc_count[$doc_id] = count($terms);

        // tf–idf, short for term frequency–inverse document frequency, 
        // according to wikipedia is a numerical statistic that is intended to reflect 
        // how important a word is to a document in a corpus

        foreach($terms as $term) {

            if(!isset($dictionary[$term])) {

                $dictionary[$term] = array('document_frequency' => 0, 'postings' => array());
            }
            if(!isset($dictionary[$term]['postings'][$doc_id])) {

                $dictionary[$term]['document_frequency']++;

                $dictionary[$term]['postings'][$doc_id] = array('term_frequency' => 0);
            }

            $dictionary[$term]['postings'][$doc_id]['term_frequency']++;
        }

        //from http://phpir.com/simple-search-the-vector-space-model/

    }

    return array('doc_count' => $doc_count, 'dictionary' => $dictionary);
}

function get_similar_documents($query='', $corpus=array(), $separator=' '){

    $similar_documents=array();

    if($query!=''&&!empty($corpus)){

        $words=explode($separator,$query);

        $corpus=get_corpus_index($corpus, $separator);

        $doc_count=count($corpus['doc_count']);

        foreach($words as $word) {

            if(isset($corpus['dictionary'][$word])){

                $entry = $corpus['dictionary'][$word];


                foreach($entry['postings'] as $doc_id => $posting) {

                    //get term frequency–inverse document frequency
                    $score=$posting['term_frequency'] * log($doc_count + 1 / $entry['document_frequency'] + 1, 2);

                    if(isset($similar_documents[$doc_id])){

                        $similar_documents[$doc_id]+=$score;

                    }
                    else{

                        $similar_documents[$doc_id]=$score;

                    }
                }
            }
        }

        // length normalise
        foreach($similar_documents as $doc_id => $score) {

            $similar_documents[$doc_id] = $score/$corpus['doc_count'][$doc_id];

        }

        // sort from  high to low

        arsort($similar_documents);

    }   

    return $similar_documents;
}

CASE 1

$query = 'are';

$corpus = array(
    1 => 'How are you?',
);

$match_results=get_similar_documents($query,$corpus);
echo '<pre>';
    print_r($match_results);
echo '</pre>';

RESULT

Array
(
    [1] => 0.52832083357372
)

CASE 2

$query = 'are';

$corpus = array(
    1 => 'how are you today?',
    2 => 'how do you do',
    3 => 'here you are! how are you? Are we done yet?'
);

$match_results=get_similar_documents($query,$corpus);
echo '<pre>';
    print_r($match_results);
echo '</pre>';

RESULTS

Array
(
    [1] => 0.54248125036058
    [3] => 0.21699250014423
)

CASE 3

$query = 'we are done';

$corpus = array(
    1 => 'how are you today?',
    2 => 'how do you do',
    3 => 'here you are! how are you? Are we done yet?'
);

$match_results=get_similar_documents($query,$corpus);
echo '<pre>';
    print_r($match_results);
echo '</pre>';

RESULTS

Array
(
    [3] => 0.6813781191217
    [1] => 0.54248125036058
)

There are plenty of improvements to be made but the model provides a way of getting good results from natural queries, which don't have boolean operators such as strpos(), preg_match(), strstr() or stristr().

NOTA BENE

Optionally eliminating redundancy prior to search the words

  • thereby reducing index size and resulting in less storage requirement

  • less disk I/O

  • faster indexing and a consequently faster search.

1. Normalisation

  • Convert all text to lower case

2. Stopword elimination

  • Eliminate words from the text which carry no real meaning (like 'and', 'or', 'the', 'for', etc.)

3. Dictionary substitution

  • Replace words with others which have an identical or similar meaning. (ex:replace instances of 'hungrily' and 'hungry' with 'hunger')

  • Further algorithmic measures (snowball) may be performed to further reduce words to their essential meaning.

  • The replacement of colour names with their hexadecimal equivalents

  • The reduction of numeric values by reducing precision are other ways of normalising the text.

RESOURCES

Answered   2023-09-20 20:26:52

Make use of substring matching using strpos():

if (strpos($string,$stringToSearch) !== false) {
    echo 'true';
}

Answered   2023-09-20 20:26:52

If you want to avoid the "falsey" and "truthy" problem, you can use substr_count:

if (substr_count($a, 'are') > 0) {
    echo "at least one 'are' is present!";
}

It's a bit slower than strpos but it avoids the comparison problems.

Answered   2023-09-20 20:26:52

  • It returns false for "are you sure?" since the position for strpos is 0 - anyone
if (preg_match('/(are)/', $a)) {
   echo 'true';
}

Answered   2023-09-20 20:26:52

  • I am getting the following warning: WARNING preg_match(): Delimiter must not be alphanumeric or backslash - anyone

Another option is to use the strstr() function. Something like:

if (strlen(strstr($haystack,$needle))>0) {
// Needle Found
}

Point to note: The strstr() function is case-sensitive. For a case-insensitive search, use the stristr() function.

Answered   2023-09-20 20:26:52

  • strstr() returns FALSE if the needle was not found. So a strlen is not necessary. - anyone

I'm a bit impressed that none of the answers here that used strpos, strstr and similar functions mentioned Multibyte String Functions yet (2015-05-08).

Basically, if you're having trouble finding words with characters specific to some languages, such as German, French, Portuguese, Spanish, etc. (e.g.: ä, é, ô, ç, º, ñ), you may want to precede the functions with mb_. Therefore, the accepted answer would use mb_strpos or mb_stripos (for case-insensitive matching) instead:

if (mb_strpos($a,'are') !== false) {
    echo 'true';
}

If you cannot guarantee that all your data is 100% in UTF-8, you may want to use the mb_ functions.

A good article to understand why is The Absolute Minimum Every Software Developer Absolutely, Positively Must Know About Unicode and Character Sets (No Excuses!) by Joel Spolsky.

Answered   2023-09-20 20:26:52

In PHP, the best way to verify if a string contains a certain substring, is to use a simple helper function like this:

function contains($haystack, $needle, $caseSensitive = false) {
    return $caseSensitive ?
            (strpos($haystack, $needle) === FALSE ? FALSE : TRUE):
            (stripos($haystack, $needle) === FALSE ? FALSE : TRUE);
}

Explanation:

  • strpos finds the position of the first occurrence of a case-sensitive substring in a string.
  • stripos finds the position of the first occurrence of a case-insensitive substring in a string.
  • myFunction($haystack, $needle) === FALSE ? FALSE : TRUE ensures that myFunction always returns a boolean and fixes unexpected behavior when the index of the substring is 0.
  • $caseSensitive ? A : B selects either strpos or stripos to do the work, depending on the value of $caseSensitive.

Output:

var_dump(contains('bare','are'));            // Outputs: bool(true)
var_dump(contains('stare', 'are'));          // Outputs: bool(true)
var_dump(contains('stare', 'Are'));          // Outputs: bool(true)
var_dump(contains('stare', 'Are', true));    // Outputs: bool(false)
var_dump(contains('hair', 'are'));           // Outputs: bool(false)
var_dump(contains('aren\'t', 'are'));        // Outputs: bool(true)
var_dump(contains('Aren\'t', 'are'));        // Outputs: bool(true)
var_dump(contains('Aren\'t', 'are', true));  // Outputs: bool(false)
var_dump(contains('aren\'t', 'Are'));        // Outputs: bool(true)
var_dump(contains('aren\'t', 'Are', true));  // Outputs: bool(false)
var_dump(contains('broad', 'are'));          // Outputs: bool(false)
var_dump(contains('border', 'are'));         // Outputs: bool(false)

Answered   2023-09-20 20:26:52

You can use the strstr function:

$haystack = "I know programming";
$needle   = "know";
$flag = strstr($haystack, $needle);

if ($flag){

    echo "true";
}

Without using an inbuilt function:

$haystack  = "hello world";
$needle = "llo";

$i = $j = 0;

while (isset($needle[$i])) {
    while (isset($haystack[$j]) && ($needle[$i] != $haystack[$j])) {
        $j++;
        $i = 0;
    }
    if (!isset($haystack[$j])) {
        break;
    }
    $i++;
    $j++;

}
if (!isset($needle[$i])) {
    echo "YES";
}
else{
    echo "NO ";
}

Answered   2023-09-20 20:26:52

The function below also works and does not depend on any other function; it uses only native PHP string manipulation. Personally, I do not recommend this, but you can see how it works:

<?php

if (!function_exists('is_str_contain')) {
  function is_str_contain($string, $keyword)
  {
    if (empty($string) || empty($keyword)) return false;
    $keyword_first_char = $keyword[0];
    $keyword_length = strlen($keyword);
    $string_length = strlen($string);

    // case 1
    if ($string_length < $keyword_length) return false;

    // case 2
    if ($string_length == $keyword_length) {
      if ($string == $keyword) return true;
      else return false;
    }

    // case 3
    if ($keyword_length == 1) {
      for ($i = 0; $i < $string_length; $i++) {

        // Check if keyword's first char == string's first char
        if ($keyword_first_char == $string[$i]) {
          return true;
        }
      }
    }

    // case 4
    if ($keyword_length > 1) {
      for ($i = 0; $i < $string_length; $i++) {
        /*
        the remaining part of the string is equal or greater than the keyword
        */
        if (($string_length + 1 - $i) >= $keyword_length) {

          // Check if keyword's first char == string's first char
          if ($keyword_first_char == $string[$i]) {
            $match = 1;
            for ($j = 1; $j < $keyword_length; $j++) {
              if (($i + $j < $string_length) && $keyword[$j] == $string[$i + $j]) {
                $match++;
              }
              else {
                return false;
              }
            }

            if ($match == $keyword_length) {
              return true;
            }

            // end if first match found
          }

          // end if remaining part
        }
        else {
          return false;
        }

        // end for loop
      }

      // end case4
    }

    return false;
  }
}

Test:

var_dump(is_str_contain("test", "t")); //true
var_dump(is_str_contain("test", "")); //false
var_dump(is_str_contain("test", "test")); //true
var_dump(is_str_contain("test", "testa")); //flase
var_dump(is_str_contain("a----z", "a")); //true
var_dump(is_str_contain("a----z", "z")); //true 
var_dump(is_str_contain("mystringss", "strings")); //true 

Answered   2023-09-20 20:26:52

  • Could you please tell me why in the world you would use a function like this, when strpos is a perfectly viable solution?... - anyone
  • @sg3s: you are totally right, however, strpos also based on something like that, also, I didn't posted it for rep just for sharing a bit of knowledge - anyone
  • last var_dump is false - anyone
  • @Sunny: it was typo: var_dump(is_str_contain("mystringss", "strings")); //true - anyone

Lot of answers that use substr_count checks if the result is >0. But since the if statement considers zero the same as false, you can avoid that check and write directly:

if (substr_count($a, 'are')) {

To check if not present, add the ! operator:

if (!substr_count($a, 'are')) {

Answered   2023-09-20 20:26:52

  • Well... partially true, in php 0 == false is true, but 0 === false is false - anyone

I had some trouble with this, and finally I chose to create my own solution. Without using regular expression engine:

function contains($text, $word)
{
    $found = false;
    $spaceArray = explode(' ', $text);

    $nonBreakingSpaceArray = explode(chr(160), $text);

    if (in_array($word, $spaceArray) ||
        in_array($word, $nonBreakingSpaceArray)
       ) {

        $found = true;
    }
    return $found;
 }

You may notice that the previous solutions are not an answer for the word being used as a prefix for another. In order to use your example:

$a = 'How are you?';
$b = "a skirt that flares from the waist";
$c = "are";

With the samples above, both $a and $b contains $c, but you may want your function to tell you that only $a contains $c.

Answered   2023-09-20 20:26:52

  • you probably meant: $found = false at the beginning - anyone
  • your function may not work if the word is linked with comma, question mark or dot. e.g. "what you see is what you get." and you want to determine if "get" is in the sentence. Notice the full stop next to "get". In this case, your function returns false. it is recommended to use regular expression or substr(I think it uses regular expression anyway) to search/replace strings. - anyone
  • @lightbringer you could not be more wrong with your recommendation, what does it mean for you "it is recommended" ? there is no supreme person that recommends or aproves. It's about the use of regular expression engine in php that is a blackhole in the language itself, you may want to try putting a regex match in a loop and benchmark the results. - anyone
  • This answer is poorly demonstrated and fails with many extended scenarios. I don't see any benefit in entertaining this technique. Here is the refined custom function and iterated call: 3v4l.org/E9dfD I have no interest in editing this wiki because I find it to be wasteful of researchers time. - anyone

Another option to finding the occurrence of a word from a string using strstr() and stristr() is like the following:

<?php
    $a = 'How are you?';
    if (strstr($a,'are'))  // Case sensitive
        echo 'true';
    if (stristr($a,'are'))  // Case insensitive
        echo 'true';
?>

Answered   2023-09-20 20:26:52

  • This is backwards. The i in stristr stands for insensitive. - anyone

It can be done in three different ways:

 $a = 'How are you?';

1- stristr()

 if (strlen(stristr($a,"are"))>0) {
    echo "true"; // are Found
 } 

2- strpos()

 if (strpos($a, "are") !== false) {
   echo "true"; // are Found
 }

3- preg_match()

 if( preg_match("are",$a) === 1) {
   echo "true"; // are Found
 }

Answered   2023-09-20 20:26:52

  • good, but preg_match is risky since it can return false or 0. You should be testing for ===1 in #3 - anyone

The short-hand version

$result = false!==strpos($a, 'are');

Answered   2023-09-20 20:26:52

  • While this code snippet may solve the question, including an explanation really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion. - anyone

Do not use preg_match() if you only want to check if one string is contained in another string. Use strpos() or strstr() instead as they will be faster. (http://in2.php.net/preg_match)

if (strpos($text, 'string_name') !== false){
   echo 'get the string';
}

Answered   2023-09-20 20:26:52

In order to find a 'word', rather than the occurrence of a series of letters that could in fact be a part of another word, the following would be a good solution.

$string = 'How are you?';
$array = explode(" ", $string);

if (in_array('are', $array) ) {
    echo 'Found the word';
}

Answered   2023-09-20 20:26:52

  • it will fail if $string is Are are, are? - anyone

You should use case Insensitive format,so if the entered value is in small or caps it wont matter.

<?php
$grass = "This is pratik joshi";
$needle = "pratik";
if (stripos($grass,$needle) !== false) { 

 /*If i EXCLUDE : !== false then if string is found at 0th location, 
   still it will say STRING NOT FOUND as it will return '0' and it      
   will goto else and will say NOT Found though it is found at 0th location.*/
    echo 'Contains word';
}else{
    echo "does NOT contain word";
}
?>

Here stripos finds needle in heystack without considering case (small/caps).

PHPCode Sample with output

Answered   2023-09-20 20:26:52

Maybe you could use something like this:

<?php
    findWord('Test all OK');

    function findWord($text) {
        if (strstr($text, 'ok')) {
            echo 'Found a word';
        }
        else
        {
            echo 'Did not find a word';
        }
    }
?>

Answered   2023-09-20 20:26:52

If you want to check if the string contains several specifics words, you can do:

$badWords = array("dette", "capitale", "rembourser", "ivoire", "mandat");

$string = "a string with the word ivoire";

$matchFound = preg_match_all("/\b(" . implode($badWords,"|") . ")\b/i", $string, $matches);

if ($matchFound) {
    echo "a bad word has been found";
}
else {
    echo "your string is okay";
}

This is useful to avoid spam when sending emails for example.

Answered   2023-09-20 20:26:52

The strpos function works fine, but if you want to do case-insensitive checking for a word in a paragraph then you can make use of the stripos function of PHP.

For example,

$result = stripos("I love PHP, I love PHP too!", "php");
if ($result === false) {
    // Word does not exist
}
else {
    // Word exists
}

Find the position of the first occurrence of a case-insensitive substring in a string.

If the word doesn't exist in the string then it will return false else it will return the position of the word.

Answered   2023-09-20 20:26:52

A string can be checked with the below function:

function either_String_existor_not($str, $character) {
    return strpos($str, $character) !== false;
}

Answered   2023-09-20 20:26:52

  • can be simplified to return strpos($str, $character) !== false - anyone

You need to use identical/not identical operators because strpos can return 0 as it's index value. If you like ternary operators, consider using the following (seems a little backwards I'll admit):

echo FALSE === strpos($a,'are') ? 'false': 'true';

Answered   2023-09-20 20:26:52

Check if string contains specific words?

This means the string has to be resolved into words (see note below).

One way to do this and to specify the separators is using preg_split (doc):

<?php

function contains_word($str, $word) {
  // split string into words
  // separators are substrings of at least one non-word character
  $arr = preg_split('/\W+/', $str, NULL, PREG_SPLIT_NO_EMPTY);

  // now the words can be examined each
  foreach ($arr as $value) {
    if ($value === $word) {
      return true;
    }
  }
  return false;
}

function test($str, $word) {
  if (contains_word($str, $word)) {
    echo "string '" . $str . "' contains word '" . $word . "'\n";
  } else {
    echo "string '" . $str . "' does not contain word '" . $word . "'\n" ;
  }
}

$a = 'How are you?';

test($a, 'are');
test($a, 'ar');
test($a, 'hare');

?>

A run gives

$ php -f test.php                   
string 'How are you?' contains word 'are' 
string 'How are you?' does not contain word 'ar'
string 'How are you?' does not contain word 'hare'

Note: Here we do not mean word for every sequence of symbols.

A practical definition of word is in the sense the PCRE regular expression engine, where words are substrings consisting of word characters only, being separated by non-word characters.

A "word" character is any letter or digit or the underscore character, that is, any character which can be part of a Perl " word ". The definition of letters and digits is controlled by PCRE's character tables, and may vary if locale-specific matching is taking place (..)

Answered   2023-09-20 20:26:52