I think the random number query will work but I have a problem with the
context of the echo line.

Here's how I get the image path.

$root_path = "/www/special_projects/Elkhart";
$agent_url = "$root_path/$agent_name";

<img src="<? echo $agent_url; ?>">

How would I use the rand function in the above statement?


> When you create the src tag in for the image, add a random number query 
> string - that will cause the browser to request the image instead of 
> using the cached version. I.e...
> echo '<img src="/images/image.jpg?'.rand(1000,9999).'">';
> -- 
> Stuart

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