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Home/ Questions/Q 8374101
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Editorial Team
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Editorial Team
Asked: June 9, 20262026-06-09T14:50:46+00:00 2026-06-09T14:50:46+00:00

I have a HTML structure: <div class=ttl> <div class=ttlpadding> <div class=item></div> </div> </div> My

  • 0

I have a HTML structure:

<div class="ttl">
   <div class="ttlpadding">
      <div class="item"></div>
   </div>
</div>

My aim is to display 10 images inside .item which will be contained in .ttlpadding which will be contained in .ttl.
So, there will be individual .ttl for each individual .item and its corresponding individual image.

My JQuery code:

    for (var i = 0; i < 10; i++) {
       $(".item").append("<a href='" + data.data[i].images.standard_resolution.url +"' ><img src='" + data.data[i].images.thumbnail.url +"' /></a>");
}

Currently, all the the ten images are getting contained inside a single “.item” and hence a single “.ttl”. I’ve tried many variations but can’t figure out the correct way.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-09T14:50:47+00:00Added an answer on June 9, 2026 at 2:50 pm
    $('.ttl').first().parent().append(
    "<div class="ttl">
       <div class="ttlpadding">
          <div class="item">" + 
    
    "<a href='" + data.data[i].images.standard_resolution.url +"' ><img src='" + data.data[i].images.thumbnail.url +"' /></a>" +
    "</div>
       </div>
    </div>"
    ); 
    

    This selects .ttl (the first one), goes to it’s parent, and adds the necessary divs and image code.

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