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timlondon

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Guys, I'm creating a javascript slideshow which works perfectly if the jpgs are prescribed as below.

 

<script type="text/javascript">

<!--

var image1=new Image()

image1.src="florida1.jpg"

var image2=new Image()

image2.src="florida2.jpg"

var image3=new Image()

image3.src="florida3.jpg"

//-->

</script>

 

but instead of using the 3 prescribed florida jpgs I wanted to echo the jpg name as php variables what's the syntax??

 

My coding is currently:

 

$xx=0;

$xxx=1;

while($xx < $totalslide) {

$preload = $preload."var image".$xxx."=new Image()";

$preload = $preload." image".$xxx.".src=".$slideimage[$xx]." ";

$xxx++;

$xx++;

}

 

When echoed, $preload produces:

 

var image1=new Image() image1.src=members/a/alpha01/Animpub/pubanalpha0118117.jpg var image2=new Image() image2.src=members/a/alpha01/Animpub/pubanalpha0118128.jpg var image3=new Image() image3.src=members/a/alpha01/Animpub/pubanalpha0118151.jpg

 

This doesn't work with the remainder of the slideshow script presumably because the jpg path isn't wrapped in double quotes. So how can I get the double quotes into the script???????

 

 

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You can reduce the code in the loop a little by doing:

<?php
$preload = array();
for($xx = 0;$xx < $totalslide; $xx++) {
   $xxx = $xx +1;
   $preload[] = 'var image'.$xxx.'=new Image()';
   $preload[] = 'image' . $xxx . '.src="' . $slideimage[$xx] . '";';
}
echo implode("\r\n",$preload)."\r\n";
?>

 

Ken

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