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Asked: 2026-05-11T01:39:52+00:00

I have written jQuery code, in files Main.html and ajax.php . The ajax.php file

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I have written jQuery code, in files Main.html and ajax.php. The ajax.php file returns the link of images to Main.html.

Now in Main.html, I have Image1, Image2, Image3, etc.

My Main.html file:

<html>     ...     # ajax.php Call     ...     # Return fields from Ajax.php </html>

My ajax.php file

echo '<a href='src1'><img src='src_path1' id='fid1' alt='Name1' /></a>Click To View image1\n'; echo '<a href='src2'><img src='src_path2' id='fid2' alt='Name2' /></a>Click To View image2\n'; // etc.

So, after executing ajax.php, I get the image locations in Main.html.

Now, when I click the Image1 link from Main.html, that corresponding image should display in the same window.

So I thought about whether again to use jQuery to view an image on the same page. How can I achieve this?

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1 Answer

  1. File ajax.php output must return the below HTML:

    <a href='#' class='imageLink' title='fid1'><img src='src1' id='fid1' alt='Name1' style='display:none;' /><span>Click To View image1</span></a>  <a href='#' class='imageLink' title='fid2'><img src='src2' id='fid2' alt='Name2' style='display:none;' /><span>Click To View image2</span></a>  <a href='#' class='imageLink' title='fid3'><img src='src3' id='fid3' alt='Name3' style='display:none;' /><span>Click To View image3</span></a>

    And jQuery code:

    $(document).ready(function(){     $('a.imageLink').click(function(){             $('#'+$(this).attr('title')).show();         $(this).find('span').hide();     }); });
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