TalkPHP - View Single Post

jaswinder_rana · 07-17-2005, 01:47 AM

one more suggestion,
you can just put all the iamge names in array and then it can spare you one more step, which is a big switch statement. like this

PHP Code:


			
<?php

$imageArr = array('ban1.gif','ban2.jpg','ban3.gif');

$iamgeCount = count($imageArr);

srand(time());

$rand = rand(0,($imageCount-1));//0-2 in this case

echo '<img src = "'.$imageArr[$rand].'" width="150" height="150"  />';

?>

and to make it more easy, you can make it read adirectory and store all the image names in an array and then use that array,

this will give you flexibility like,
in the first method, you can put any number of images in array
in second, just put images in directory and it'll automatically pick them up.

and ofcourse, using database is a totally different option

hope this helps

07-17-2005, 01:47 AM	#8 (permalink)
jaswinder_rana The Acquainted Join Date: May 2005 Posts: 106 Thanks: 0	one more suggestion, you can just put all the iamge names in array and then it can spare you one more step, which is a big switch statement. like this PHP Code: `<?php $imageArr = array('ban1.gif','ban2.jpg','ban3.gif'); $iamgeCount = count($imageArr); srand(time()); $rand = rand(0,($imageCount-1));//0-2 in this case echo '<img src = "'.$imageArr[$rand].'" width="150" height="150" />'; ?>` and to make it more easy, you can make it read adirectory and store all the image names in an array and then use that array, this will give you flexibility like, in the first method, you can put any number of images in array in second, just put images in directory and it'll automatically pick them up. and ofcourse, using database is a totally different option hope this helps __________________ --------------------------- Errors = Improved Programming. Portfolio