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Posted by guywmustang on 12/28/06 03:30
Easiest way would just be to at the beginning of image.php to add the
following line at the top of the code:
<?
require_once("first_view.php");
.....
Then create a function in there
function GetTitle( $ImageNumber )
{
return m_The_Image_Title;
}
and maybe another function which gives you the image source code?
function GetImage( $ImageNumber )
{
echo ' <img src="'.$imgNum.'" width="200" border="1"
alt="'.GetTitle($ImageNumber).'"> ';
}
That'd be easiest seeing as how it appears you have access to the
first_view.php file, so you can access methods in there.
Jason
On Dec 27, 8:20 pm, "eholz1" <ewh...@gmail.com> wrote:
> Hello PHP Group,
>
> I have a php page that calls another php page to load a image from a
> database (mysql) in to the calling page.
> It works like this:
> first_view.php has
> <img src="image.php?img=5" width="200" border="1" alt="Image of
> Sunset">
> the image.php opens the database and returns the image to the page
> (first_view.php).
>
> image.php can (does) also pull the title of the image being loaded to
> first_view.php out of the query where the image data is send back:
>
> I send the image data back like this:
>
> //echo back the image data
> header("Content-type: image/jpeg");
> echo base64_decode($encodeddata);
>
> I have a <p></p> tag under the image tag, and would like to put the
> title from the database there.
> How can I pass the title of the image back to first_view from
> image.php.
>
> There is no html in image.php.
> Thanks for a good group!
>
> ewholz
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