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Home/ Questions/Q 8792777
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T23:01:43+00:00 2026-06-13T23:01:43+00:00

I have the following code <a href=# class=chosenflag data-flag=<?php echo $lang;?>><img class=flag src=/img/flag/United Kingdom.png

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I have the following code

<a href="#" class="chosenflag" data-flag="<?php echo $lang;?>"><img class="flag" src="/img/flag/United Kingdom.png" alt="" /></a>

<ul class="choices">
<li><a href="#" data-flag="uk"><img class="flag" src="/img/flag/United Kingdom.png" alt="" /></a></li>
<li><a href="#" data-flag="de"><img class="flag" src="/img/flag/Germany.png" alt="" /></a></li>
<li><a href="#" data-flag="es"><img class="flag" src="/img/flag/Spain.png" alt="" /></a></li>
<li><a href="#" data-flag="fr"><img class="flag" src="/img/flag/France.png" alt="" /></a></li>
<li><a href="#" data-flag="it"><img class="flag" src="/img/flag/Italy.png" alt="" /></a></li>
</ul>

The link with the class ‘chosenflag’ is the active flag which picks up a php variable which will either be uk, de, es, fr, it.

What I would like to happen is somehow get the img src from the ul list to the chosen flag img src depending on the data-flag attribute.

Would someone be able to help me out?

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  1. Editorial Team
    Editorial Team
    2026-06-13T23:01:44+00:00Added an answer on June 13, 2026 at 11:01 pm

    This should work if you click on one of the flags. It should put its flag image in the chosen flag image spot.

    $chosen = $('.chosenflag')
    $('.choices').on('click', 'a', function () {
      // changes data-flag on .chosenflag
      var dataFlag = this.data('flag');
      $chosen.data('flag', dataFlag);
      // 'this' is the clicked on <a>
      var src = this.find('img').attr('src');
      $chosenImg = $chosen.find('img');
      $chosenImg.attr('src', src);
    });
    

    I think this is what you are looking for.

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