Php Image Syntax Question?

[scm22ri]

Hi Everyone,

 

I'm trying to figure out the correct syntax when it comes to displaying an image on my website.

 

When I hard code the number in it works but when I add the variable $id it's not working.

 

// Not working

 

echo '<div align="center"><img src="photos'. $id . rand(1,3) . '.jpg" width="250" alt="Random Image" /></div></td>';
echo '<div align="center"><img src="photos'. $id .''. rand(1,3) . '.jpg" width="250" alt="Random Image"/></div></td>';
echo '<div align="center"><img src="photos'.$id. rand(1,3) . '.jpg" width="250" alt="Random Image" /></div></td>';

echo '<div align="center"><img src="photos/'.$id.'/'.echo rand(1,n);'.jpg" width="250" /></div></td>';
echo '<div align="center"><img src="photos/'. $id . rand(1,3) . '.jpg" width="250" alt="Random Image" /></div></td>';

 

//Works but only when I hard code the number 4 (but I can't have that because every photo is in a different folder.

echo '<div align="center"><img src="photos'. '/4/' . rand(1,3) . '.jpg" width="250" alt="Random Image" /></div></td>';

 

Any help would be appreciated, thanks!

 

Hey Everyone,

 

I finally solved the problem. The correct syntax is

echo '<div align="center"><img src="photos/'. $id .'/'. rand(1,3) . '.jpg" width="250" alt="Random Image"/></div></td>';

Edited December 7, 2012 by scm22ri