Need a script to tell me where a hyperlink originated

[Dynamo]

Not sure if this is a php or an html question but I'll post it anyway.

I have multiple thumbnail images on my catalogue page catalogue.php and they are
all hyperlinked to a single page called desc.php that will give an enhanced
description of the product selected.

I want the desc.php page to print the contents of a text file named ?.txt based
on which image was clicked on in my catalogue page.

example: if I click on 1.jpg in my catalogue then I want the file 1.txt to be
selected and printed on my description page.

I know how to print the contents of a textfile and I know how to disect the
strings of the filenames so that I can associate 1.txt with 1.jpg.

What I dont know is how to retrieve the name of the image file that was clicked
on the catalogue page within my script for the desc.php page

If I didn't use a hyperlink image and used a form submit button instead on my
catalogue page then of course I could use something like:

$filename = $_POST['filename']

Hope you understood all of that.

Regards
Dynamo

[kingofkolt]

"Dynamo" <Dy***********@newsguy.com> wrote in message
news:co*********@drn.newsguy.com...

Not sure if this is a php or an html question but I'll post it anyway.

I have multiple thumbnail images on my catalogue page catalogue.php and they are all hyperlinked to a single page called desc.php that will give an enhanced description of the product selected.

I want the desc.php page to print the contents of a text file named ?.txt based on which image was clicked on in my catalogue page.

example: if I click on 1.jpg in my catalogue then I want the file 1.txt to be selected and printed on my description page.

I know how to print the contents of a textfile and I know how to disect the strings of the filenames so that I can associate 1.txt with 1.jpg.

What I dont know is how to retrieve the name of the image file that was clicked on the catalogue page within my script for the desc.php page

If I didn't use a hyperlink image and used a form submit button instead on my catalogue page then of course I could use something like:

$filename = $_POST['filename']

Hope you understood all of that.

Regards
Dynamo

The easiest and most common method is to use $_GET to indicate the filename.
Other than that, I believe the only way to use an image the way you want to
use it, is to use it as a form submit button.

- JP

[Gary L. Burnore]

On Mon, 29 Nov 2004 16:09:47 GMT, "kingofkolt"
<je**********@comcast.net> wrote:

"Dynamo" <Dy***********@newsguy.com> wrote in message
news:co*********@drn.newsguy.com...

Not sure if this is a php or an html question but I'll post it anyway.

I have multiple thumbnail images on my catalogue page catalogue.php and

they are

all hyperlinked to a single page called desc.php that will give an

enhanced

description of the product selected.

I want the desc.php page to print the contents of a text file named ?.txt

based

on which image was clicked on in my catalogue page.

example: if I click on 1.jpg in my catalogue then I want the file 1.txt to

be

selected and printed on my description page.

I know how to print the contents of a textfile and I know how to disect

the

strings of the filenames so that I can associate 1.txt with 1.jpg.

What I dont know is how to retrieve the name of the image file that was

clicked

on the catalogue page within my script for the desc.php page

If I didn't use a hyperlink image and used a form submit button instead on

my

catalogue page then of course I could use something like:

$filename = $_POST['filename']

Hope you understood all of that.

Regards
Dynamo

The easiest and most common method is to use $_GET to indicate the filename.
Other than that, I believe the only way to use an image the way you want to
use it, is to use it as a form submit button.

Depending on the code you use to load the image and build the
hyperlink around it, you'd simply add the name of the file to each
hyperlink as you build it:

<A HREF="desc.php?selected=1.jpg" SRC="1.jpg">
<A HREF="desc.php?selected=2.jpg" SRC="2.jpg">
for example.
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[Ken Robinson]

Dynamo wrote (in part):

I have multiple thumbnail images on my catalogue page catalogue.php and they are all hyperlinked to a single page called desc.php that will give an enhanced description of the product selected.

I want the desc.php page to print the contents of a text file named ?.txt based on which image was clicked on in my catalogue page.

example: if I click on 1.jpg in my catalogue then I want the file 1.txt to be selected and printed on my description page.

How are you specifying the link now?

If you have something like this:
<a href="desc.php"><img src=1.jpg></a>

You should do:
<a href="desc.php?img=1"><img src=1.jpg></a>

In your file desc.php, you obtain the value by looking at $_GET['img']
Ken

[Dynamo]

In article <11**********************@f14g2000cwb.googlegroups .com>, Ken Robinson
says...

How are you specifying the link now?

If you have something like this:
<a href="desc.php"><img src=1.jpg></a>

You should do:
<a href="desc.php?img=1"><img src=1.jpg></a>

In your file desc.php, you obtain the value by looking at $_GET['img']
Ken

Thank you so much!! That seems to work. Don't even have to use the form tag in
my html. In addition, by assigning the value 1 to img I dont have to worry about
disecting the strings to associate 1.jpg with 1.txt. I can use something like
$filetoopen= $_GET['img']. '.txt'

Many thanks
Dynamo