RE: [PHP] Javascript and php

[Reinhart Viane]

Thank you very much,

After some small adjustements to make it work with the rest of the
javascript it works like a charm!!

Have a nice Sunday :)

Greetings,
Reinhart

-----Original Message-----
From: Bruno B B Magalhães [mailto:[EMAIL PROTECTED] 
Sent: zondag 7 november 2004 12:33
To: [EMAIL PROTECTED]
Subject: Re: [PHP] Javascript and php


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
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