jQuery how to find an element based on a data-attribute value?

1,469,501

Solution 1

You have to inject the value of current into an Attribute Equals selector:

$("ul").find(`[data-slide='${current}']`)

For older JavaScript environments (ES5 and earlier):

$("ul").find("[data-slide='" + current + "']"); 

Solution 2

in case you don't want to type all that, here's a shorter way to query by data attribute:

$("ul[data-slide='" + current +"']");

FYI: http://james.padolsey.com/javascript/a-better-data-selector-for-jquery/

Solution 3

When searching with [data-x=...], watch out, it doesn't work with jQuery.data(..) setter:

$('<b data-x="1">'  ).is('[data-x=1]') // this works
> true

$('<b>').data('x', 1).is('[data-x=1]') // this doesn't
> false

$('<b>').attr('data-x', 1).is('[data-x=1]') // this is the workaround
> true

You can use this instead:

$.fn.filterByData = function(prop, val) {
    return this.filter(
        function() { return $(this).data(prop)==val; }
    );
}

$('<b>').data('x', 1).filterByData('x', 1).length
> 1

Solution 4

Without JQuery, ES6

document.querySelectorAll(`[data-slide='${current}']`);

I know the question is about JQuery, but readers may want a pure JS method.

Solution 5

I improved upon psycho brm's filterByData extension to jQuery.

Where the former extension searched on a key-value pair, with this extension you can additionally search for the presence of a data attribute, irrespective of its value.

(function ($) {

    $.fn.filterByData = function (prop, val) {
        var $self = this;
        if (typeof val === 'undefined') {
            return $self.filter(
                function () { return typeof $(this).data(prop) !== 'undefined'; }
            );
        }
        return $self.filter(
            function () { return $(this).data(prop) == val; }
        );
    };

})(window.jQuery);

Usage:

$('<b>').data('x', 1).filterByData('x', 1).length    // output: 1
$('<b>').data('x', 1).filterByData('x').length       // output: 1

// test data
function extractData() {
  log('data-prop=val ...... ' + $('div').filterByData('prop', 'val').length);
  log('data-prop .......... ' + $('div').filterByData('prop').length);
  log('data-random ........ ' + $('div').filterByData('random').length);
  log('data-test .......... ' + $('div').filterByData('test').length);
  log('data-test=anyval ... ' + $('div').filterByData('test', 'anyval').length);
}

$(document).ready(function() {
  $('#b5').data('test', 'anyval');
});

// the actual extension
(function($) {

  $.fn.filterByData = function(prop, val) {
    var $self = this;
    if (typeof val === 'undefined') {
      return $self.filter(

        function() {
          return typeof $(this).data(prop) !== 'undefined';
        });
    }
    return $self.filter(

      function() {
        return $(this).data(prop) == val;
      });
  };

})(window.jQuery);


//just to quickly log
function log(txt) {
  if (window.console && console.log) {
    console.log(txt);
    //} else {
    //  alert('You need a console to check the results');
  }
  $("#result").append(txt + "<br />");
}
#bPratik {
  font-family: monospace;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

<div id="bPratik">
  <h2>Setup</h2>
  <div id="b1" data-prop="val">Data added inline :: data-prop="val"</div>
  <div id="b2" data-prop="val">Data added inline :: data-prop="val"</div>
  <div id="b3" data-prop="diffval">Data added inline :: data-prop="diffval"</div>
  <div id="b4" data-test="val">Data added inline :: data-test="val"</div>
  <div id="b5">Data will be added via jQuery</div>
  <h2>Output</h2>
  <div id="result"></div>

  <hr />
  <button onclick="extractData()">Reveal</button>
</div>

Or the fiddle: http://jsfiddle.net/PTqmE/46/

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Jannis
Author by

Jannis

Updated on January 15, 2021

Comments

  • Jannis
    Jannis over 3 years

    I've got the following scenario:

    var el = 'li';
    

    and there are 5 <li>'s on the page each with a data-slide=number attribute (number being 1,2,3,4,5 respectively).

    I now need to find the currently active slide number which is mapped to var current = $('ul').data(current); and is updated on each slide change.

    So far my tries have been unsuccessful, trying to construct the selector that would match the current slide:

    $('ul').find(el+[data-slide=+current+]);
    

    does not match/return anything…

    The reason I can't hardcode the li part is that this is a user accessible variable that can be changed to a different element if required, so it may not always be an li.

    Any ideas on what I'm missing?

    • xandy
      xandy over 13 years
      you sure within your .find(el+[data-slide=+current+]); is the code that you write? it seems you missed some quotations to "[data-slide]"
    • Avatar
      Avatar over 4 years
      That's what helped me to select all data attributes (regardless the value): $('*[data-slide]') You can use it with e.g. $('*[data-slide]').each( function() { ... });
  • AaronLS
    AaronLS over 11 years
    I worry that this won't work if data was added via jQuery .data(...) function, since this doesn't always render an attribute and soemtimes uses browser specific storage mechanism. Another reason that I wish jQuery core had a data specific selector.
  • Pawel
    Pawel over 10 years
    Notice that it doesn't work for elements where you set data with $('#element').data('some-att-name', value); but only for those with hardcoded attribute. I've got this problem and to make it work set data by writing the attribute directly $('#element').attr('data-some-att-name', value);
  • akousmata
    akousmata over 9 years
    For those of you who care, given the structure <ul><li data-slide="item"></li></ul> this answer's selector will return undefined so it does not work given the user's scenario. This answer WILL work for the structure <ul data-slide="item"><li></li></ul> While I understand why its popularity due to brevity, it is technically an incorrect answer. jsfiddle.net/py5p2abL/1
  • Artem
    Artem over 2 years
    Just put a space between the ul and the brackets and it will work.