HTTP_REFERER, or am I stupid

[dcjones]

Hi all.

Problem: I have a script that calls another script which runs a query. How do I get the second script retrun to the first script when the user clicks on the back button included on the page..

I am trying to use HTTP_REFERER without much luck. This is what I have:

BUTTON on SECOND PAGE

<tr>
<td><a href="\" .$HTTP_REFERER. "\" onMouseOut="MM_swapImgRestore()" onMouseOver="MM_swapImage('handl','','../../Gif/HandR.gif',1)" target="_self"><img name="handl" border="0" src="../../Gif/HandG.gif" width="35" height="12"></a>
<font color="#990000" size="-1">Back</font></td>
</tr>

This is a button which I want to use to take the user back to the calling (FIRST ) script.

The link from the FIRST script ito the SECOND script using .$row["name"] for the query.

<a href=\"accomodation/php/ac.php?idname=<? echo urlencode($row["name"]).">".$row["name"] ?></a>

The scripts work OK apart form the back button.

Am I barking up the wrong tree and should I be doing this another way.

I know that this is not very clear but I hope you can see what I am trying to do.

Thanks in advance

DCJ

[superwormy]

That really should work... although you may have to use $_SERVER['HTTP_REFERER'] instead of jsut $HTTP_REFERER

What exactly is wrong with it... what does $HTTP_REFERER print?

Oh and wait... you are using PHP to PRINT that line right? ie:

print ("<a href=\" . $_SERVER[HTTP_REFERER] . " onMouseOut...");

Otherwise, you'd need to at least put the variable in a PHP block and print that:

<a href="<?php print ($_SERVER[HTTP_REFERER]); ?>" onMouseOut="MM_swapImgRestore()" onMouseOver="MM_swapImage('hand...

[dcjones]

Hi there Superwormy,

You have the answer, thanks I used

<a href="<?php print ($_SERVER[HTTP_REFERER]); ?>" onMouseOut="MM_swapImgRestore()" onMouseOver="MM_swapImage

All problems fixed,

Many thanks

Dereck