[SOLVED] Really quick question

[tfburges]

What's wrong with this?

 

A field has a hyperlink beside it with the URL linked to the field description so that the field description and any matching data within the database pertaining to that field can be queried.

 

Example URL (all variables exist):

saac.php?fnvs=",$selected_formx_v,"&qf=",$dfcdi,"

 

Then in saac.php...

$qf = $_GET['qf'];
$qpf = $_POST[$qf];
echo $qpf;

 

$qf works fine but $qpf doesn't.  It must be a syntax error or something special that I'm not aware of.

 

I've also tried $qpf = $_POST['$qf']; and other ways to put single quotes around $qf.

[JD*]

$_POST only works when you post the infomation, i.e. you submit the form. So unless I misread, you're clicking on a hyperlink to get a description, but not submitting the form and that's where the problem is.

[MadTechie]

why would  $_POST[$qf] work ?

its a get not a post !

 

[New Coder]

With

 

$qf = $_GET['qf'];

$qpf = $_POST[$qf];

echo $qpf;

 

it looks like you are trying to put the variable $gf that you have recieved via the url into a new variable $gpf... if so

 

$qpf = $qf;

 

will work. if your trying to then send the variable $gpf somewhere else via post I would put it into a hidden input type.

 

I may have totally misunderstood what you are doing

post all your code

[tfburges]

$_POST only works when you post the infomation, i.e. you submit the form. So unless I misread, you're clicking on a hyperlink to get a description, but not submitting the form and that's where the problem is.

 

OHHHHHHHHHH LOLOLOLOLOL WOW I CAN'T BELIEVE I OVERLOOKED THAT!

 

THANKS!!1!11ONE

[JD*]

Don't forget to mark your post as solved if it's fixed.