Re: Javascript and php

[Bruno B B Magalhães <brunobbm@xxxxxxxxxx>]

Reinhart,


<?php
for($i = 0; $i < mysql_num_rows($result)-1; $i++)
{
	$row = mysql_fetch_object($result);
	
	echo '"'.$row->picture_url.'"';
	
	if($i <= mysql_num_rows($result)-2)
	{
		echo ',';
	}
}
?>

Here is how I do in my developments.

Regards,
Bruno B B Magalhães

On Nov 7, 2004, at 8:44 AM, Reinhart Viane wrote:

> Hey all,
> Hope some of you also work on sundays :)
>
> I have a little javascript which displays a images (with previous / 
> next
> thing)
> Now, i populate the javascript array with an php array:
>
> <SCRIPT LANGUAGE="JavaScript">
>
> <!-- Begin
> NewImg = new Array (
> <?php
> while($row = mysql_fetch_object($result)){
> echo "\"".$row->picture_url."\",";
> }
> ?>
> "../pictures/7_stripper3.jpg",
> "../pictures/7_stripper2.jpg"
> );
> var ImgNum = 0;
> var ImgLength = NewImg.length - 1;
> ...
> </script>
>
> As you can see i echo the url of the picture.
> After each picture url there needs to be a ','
> But not after the last picture.
> At this moment all pictures have the ',' after there url, even the last
> one.
> Any way to determine if the url is the url of the last picture and thus
> not printing a ',' behind that last one?
>
> Thx in advance,
>
> Reinhart
>
>   _____
>
> Reinhart Viane
>  <mailto:rv@xxxxxxxx> rv@xxxxxxxx
> Domos || D-Studio
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