Is there a way to do this?

[sks1024]

If have a page that reads from a db and gets the values of the fields. If one of the fields in the table is not NULL then I call a page in an image tag (pic.php) that that displays an image that I added text on top of with gd. The only problem I have is how to I get the output of the query to that other page (pic.php). The query outputs the name of the image that is stored in the db. i.e. image.jpg That has to go to the pic.php page. It works great when I hardcode the name of the image in the pic.php page but I want to pass it the output of the db field to it.

Is there a way to do this.

Here is the line that needs the db output page named pic.php

$source = ImageCreateFromJPEG("/home/u1/stansawyer/html/upload_pics/$db_field_output");

Here is the image tag on the main page that calls pic.php if needed

print "<img src=pic.php>";

Any help would be thankfull.

Stan

[sarahk]

try

print "<img src='pic.php?name=$db_field_output'>";

this will get output to the browser as

<img src='pic.php?name=funnypic.jpg'>

which will pass funnypic.jpg to pic.php.

hope this helps.

[sks1024]

That works alot better than the way I was doing it. I was fopen the file and trying to replace a string with the variable.

Both ways work, yours is better and easier.

Thanks,
Stan