jquery - get array of elements with identical data attribute values

Question

Given that I have a document similar to this:

<div>
    <div data-myAttr = "foo"></div>
    <div data-myAttr = "bar"></div>
</div>
<div>
    <div data-myAttr = "foo"></div>
    <div data-myAttr = "bar"></div>
</div>

How can I get all elements that contain data-myAttr with foo's and bar's grouped?

Edit: this is more of a theoretical question. I have no tried code that has failed but am exploring a more elegant option for code that already exists.

BTW, thank you for the -1!?

Edit: can someone mark this as duplicate? Found applicable answer here: Group and count HTML elements by data attribute in jQuery

For exploring more elegant options to already working code, consider codereview.stackoverflow.com , and you'll still want to post your code. — James Montagne
– James Montagne, Commented Apr 8, 2013 at 15:48

TJ Sherrill · Accepted Answer · 2013-06-18 23:24:54Z

2

Ok this is my first ever answer on here so don't judge to harshly... In my case I had duplicates in my array as well. And I am only just starting to fully understand how this all works but here goes:

elements = $('div[data-myAttr]').map(function () {
    return $(this).attr('data-myAttr');
});

Then run a basic js loop by name to grab all elements with the correct attr.

answered Jun 18, 2013 at 23:24

TJ Sherrill

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Comments

Adil · Accepted Answer · 2013-04-08 15:55:31Z

0

you can use attribute selector and combine two attribute selectors with comma;

Live Demo

$('[data-myAttr=foo], [data-myAttr=bar]')

You can iterate through the elements using each()

$('[data-myAttr=foo], [data-myAttr=bar]').each(function(){
  alert($(this).text());  
});

edited Apr 8, 2013 at 15:55

answered Apr 8, 2013 at 15:39

Adil

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3 Comments

Chris Stahl Over a year ago

Thanks for the answer but, is it possible to do this without having to define the data-myAttr value?

Chris Stahl Over a year ago

Sorry, I meant without having to define it as part of the selector. Thank you, though!

Adil Over a year ago

Can you explain "without having to define it as part of the selector"?

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