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[SOLVED] Show last row - code to display an image


Bricktop

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Hello all,

 

I am using the below code to display a list of thumnail images, then I'm using some Javascript code to display the full-size image using an "onclick" event.

 

All of this works tickety boo, but when the gallery page loads, because no one has yet clicked a thumbnail, the main image holder on the page just shows a blank box with an ugly image not found red cross.

 

What I would like to do is somehow always show the first image in the set.  Is there anyway I can achieve this do you know?  I hope so but I'm a little concerned as to the fact that it's an onclick event I'm using.

 

My code is below - I've added a firsrow=true/false line into the loop already - maybe we could work on that?

 

Thanks

 


    	$sql = mysql_query("SELECT * FROM gallery ORDER BY id");
	$firstRow = true;
	while ($a = mysql_fetch_array($sql)) 
	{
$content .= '<li><a href="'.$a['path'].''.$a['filename'].'" onclick="showPicture(this); return false;" title="'.str_replace('"','"',$a[description]).'"><img src="'.$a['thumbnailpath'].''.$a['filename'].'" alt="'.str_replace('"','"',$a[description]).'" /></a></li>';
$firstRow = false;
    		}

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You could use something like this:

 

<script>

  function fixBrokenImage(){ 
  

    var imglst = document.images; 
    for(var i = 0; i < imglst.length; i++){ 
      imglst[i].onerror = function() { 
        this.src = "myimage.gif"; 
      } 
      imglst[i].src = imglst[i].src; 
    } 
  } 

</script>

 

then use <body onLoad="fixBrokenImage();">

 

Just change myimage to the image you want to use as default.

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Hi Swamp,

 

Thanks but just cracked it - the HTML code to show dynamically show the image was:

 

<img id="placeholder" src="" alt="Image Viewer" />

 

I assumed the "src=" had to be kept blank - I just tried inserting an image URL into that and the image still changes with the onlick event!

 

Thanks for looking into it for me though matey!

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