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beginning_php thread: RE: Let's try another tact: passing variable url into a new window.


Message #1 by Jefferis Peterson <jefferis@p...> on Mon, 30 Sep 2002 19:04:55 -0400
2 Problems so far:
<a href='<?php echo "show_image.php?img={$row_viewall['pictures']}" ;?>'
target ="new"> LargePictures</a></font>

But the image source is blank on the new page:
> <IMG src="" />
> </DIV>

And the page is a full page rather than one of limited dimensions. I can
probably modify this with a javascript onWindow open action.

The other try:
<?php echo "<a href=\"show_image.php?img={$row_viewall['pictures']}\">";
echo "LargePictures </a>";?>
Also produces a blank image, so somehow the variable value is  not being
transmitted BUT the link is showing the proper id info in the link area of
the browser page:
http://www.yellow-diamonds.com/show_image.php?img=images/B-06A-02.jpg

Opens the pageOn 9/30/02 5:45 PM, "Nikolai Devereaux" <yomama@u...>
wrote:

> 
> Jeff,
> 
> Take a closer look at what you're outputting!!  Do you see your mistake?
> 
> 
> <quote>
> 
>> From this code:
> 
> <a href='<?php echo "<a
> href=\"show_image.php?img={$row_viewall['pictures']}\">";?>'>LargePictures</a>
> 
> </quote>
> 
> You're creating a link to "<a href=...".  If you view the source of your
> generated page, you'll see something like this:
> 
> <a href='<a href="...">'>
> 
> you have two options.
> 
> 1)  remove the <a href=\" stuff from the PHP code.
> 2)  remove the <a href=' stuff from the HTML code.
> 
> 
> nik
> 
> 

~~~~~~~~~~~~
Jefferis Peterson, Pres.
Web Design and Marketing
http://www.PetersonSales.net
Tel .  xxx-xxx-xxxx
ICQ 19112253

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