How to get the children of the $(this) selector?

Asked 2023-09-20 20:31:19 View 370,238

I have a layout similar to this:

<div id="..."><img src="..."></div>

and would like to use a jQuery selector to select the child img inside the div on click.

To get the div, I've got this selector:

$(this)

How can I get the child img using a selector?

Answers

The jQuery constructor accepts a 2nd parameter called context which can be used to override the context of the selection.

jQuery("img", this);

Which is the same as using .find() like this:

jQuery(this).find("img");

If the imgs you desire are only direct descendants of the clicked element, you can also use .children():

jQuery(this).children("img");

Answered   2023-09-20 20:31:19

  • for what it's worth: jQuery("img", this) is internally converted into: jQuery(this).find("img") So the second is ever so slightly faster. :) - anyone

You could also use

$(this).find('img');

which would return all imgs that are descendants of the div

Answered   2023-09-20 20:31:19

  • In general cases, it seems like $(this).children('img') would be better. e.g. <div><img src="..." /><div><img src="..." /></div></div> because presumably the user wants to find 1st-level imgs. - anyone

If you need to get the first img that's down exactly one level, you can do

$(this).children("img:first")

Answered   2023-09-20 20:31:19

If your DIV tag is immediately followed by the IMG tag, you can also use:

$(this).next();

Answered   2023-09-20 20:31:19

  • .next() is really fragile, unless you can always guarantee the element will be the next element, it's better to use a different method. In this case, the IMG is a descendant, not a sibling, of the div. - anyone

The direct children is

$('> .child-class', this)

Answered   2023-09-20 20:31:19

You can find all img element of parent div like below

$(this).find('img') or $(this).children('img')

If you want a specific img element you can write like this

$(this).children('img:nth(n)')  
// where n is the child place in parent list start from 0 onwards

Your div contains only one img element. So for this below is right

 $(this).find("img").attr("alt")
                  OR
  $(this).children("img").attr("alt")

But if your div contain more img element like below

<div class="mydiv">
    <img src="test.png" alt="3">
    <img src="test.png" alt="4">
</div>

then you can't use upper code to find alt value of second img element. So you can try this:

 $(this).find("img:last-child").attr("alt")
                   OR
 $(this).children("img:last-child").attr("alt")

This example shows a general idea that how you can find actual objects within the parent object. You can use classes to differentiate your child's object. That is easy and fun. i.e.

<div class="mydiv">
    <img class='first' src="test.png" alt="3">
    <img class='second' src="test.png" alt="4">
</div>

You can do this as below :

 $(this).find(".first").attr("alt")

and more specific as:

 $(this).find("img.first").attr("alt")

You can use find or children as above code. For more visit Children http://api.jquery.com/children/ and Find http://api.jquery.com/find/. See example http://jsfiddle.net/lalitjs/Nx8a6/

Answered   2023-09-20 20:31:19

Ways to refer to a child in jQuery. I summarized it in the following jQuery:

$(this).find("img"); // any img tag child or grandchild etc...   
$(this).children("img"); //any img tag child that is direct descendant 
$(this).find("img:first") //any img tag first child or first grandchild etc...
$(this).children("img:first") //the first img tag  child that is direct descendant 
$(this).children("img:nth-child(1)") //the img is first direct descendant child
$(this).next(); //the img is first direct descendant child

Answered   2023-09-20 20:31:19

Without knowing the ID of the DIV I think you could select the IMG like this:

$("#"+$(this).attr("id")+" img:first")

Answered   2023-09-20 20:31:19

  • this probably actually works but it's kinda the Rube Goldberg answer :) - anyone

Try this code:

$(this).children()[0]

Answered   2023-09-20 20:31:19

  • @rémy: That's not even valid syntax. (Took all of 2 years for anyone to notice...) - anyone

You can use either of the following methods:

1 find():

$(this).find('img');

2 children():

$(this).children('img');

Answered   2023-09-20 20:31:19

jQuery's each is one option:

<div id="test">
    <img src="testing.png"/>
    <img src="testing1.png"/>
</div>

$('#test img').each(function(){
    console.log($(this).attr('src'));
});

Answered   2023-09-20 20:31:19

You can use Child Selecor to reference the child elements available within the parent.

$(' > img', this).attr("src");

And the below is if you don't have reference to $(this) and you want to reference img available within a div from other function.

 $('#divid > img').attr("src");

Answered   2023-09-20 20:31:19

Also this should work:

$("#id img")

Answered   2023-09-20 20:31:19

Here's a functional code, you can run it (it's a simple demonstration).

When you click the DIV you get the image from some different methods, in this situation "this" is the DIV.

$(document).ready(function() {
  // When you click the DIV, you take it with "this"
  $('#my_div').click(function() {
    console.info('Initializing the tests..');
    console.log('Method #1: '+$(this).children('img'));
    console.log('Method #2: '+$(this).find('img'));
    // Here, i'm selecting the first ocorrence of <IMG>
    console.log('Method #3: '+$(this).find('img:eq(0)'));
  });
});
.the_div{
  background-color: yellow;
  width: 100%;
  height: 200px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<div id="my_div" class="the_div">
  <img src="...">
</div>

Hope it helps!

Answered   2023-09-20 20:31:19

You may have 0 to many <img> tags inside of your <div>.

To find an element, use a .find().

To keep your code safe, use a .each().

Using .find() and .each() together prevents null reference errors in the case of 0 <img> elements while also allowing for handling of multiple <img> elements.

// Set the click handler on your div
$("body").off("click", "#mydiv").on("click", "#mydiv", function() {

  // Find the image using.find() and .each()
  $(this).find("img").each(function() {
  
        var img = this;  // "this" is, now, scoped to the image element
        
        // Do something with the image
        $(this).animate({
          width: ($(this).width() > 100 ? 100 : $(this).width() + 100) + "px"
        }, 500);
        
  });
  
});
#mydiv {
  text-align: center;
  vertical-align: middle;
  background-color: #000000;
  cursor: pointer;
  padding: 50px;
  
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>

<div id="mydiv">
  <img src="" width="100" height="100"/>
</div>

Answered   2023-09-20 20:31:19

$(document).ready(function() {
  // When you click the DIV, you take it with "this"
  $('#my_div').click(function() {
    console.info('Initializing the tests..');
    console.log('Method #1: '+$(this).children('img'));
    console.log('Method #2: '+$(this).find('img'));
    // Here, i'm selecting the first ocorrence of <IMG>
    console.log('Method #3: '+$(this).find('img:eq(0)'));
  });
});
.the_div{
  background-color: yellow;
  width: 100%;
  height: 200px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<div id="my_div" class="the_div">
  <img src="...">
</div>

Answered   2023-09-20 20:31:19

If your img is exactly first element inside div then try

$(this.firstChild);

$( "#box" ).click( function() {
  let img = $(this.firstChild);
  console.log({img});
})
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

<div id="box"><img src="https://picsum.photos/seed/picsum/300/150"></div>

Answered   2023-09-20 20:31:19

With native javascript you can use

if you've more than one image tag then use

this.querySelectorAll("img")

if only one image tag then us

this.querySelector("img")

Answered   2023-09-20 20:31:19

You could use

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js">
 $(this).find('img');
</script>

Answered   2023-09-20 20:31:19