$_SERVER

[jrobles]

I have an include that processes the log in form for located at the header of every page. The problem is, when it comes to my contact page now I have two forms. So when i process the contact form the header form is checked because the page is sending a post. Is there away that I can use the code below but only have it look at one particular form, so when any other forms are processed it ignores them?

$_SERVER[REQUEST_METHOD]=='POST'

[Daniel0]

You could use the name of the submit button or you could add a hidden field.

 

By the way, check out the manual's section on arrays. Particularly, I'm thinking on the part where it tells how to access indices in arrays.

[jrobles]

not sure i follow your solution. How do i use the $_SERVER code with the hidden field?

[mrMarcus]

not sure i follow your solution. How do i use the $_SERVER code with the hidden field?

don't worry about the $_SERVER[REQUEST_METHOD]=='POST' .. as Daniel said, simply use unique submit names for each form:

 

login form in header

...form stuff;
<input type="submit" name="login_header" value="Login" />

 

contact form

...contact form stuff;
<input type="submit" name="contact_form" value="Contact Us" />

[mrMarcus]

and, are both forms being processed by the same include file?

[jrobles]

no, the login form is being processed by the include which does some cURL stuff. The contact form has a quick if statement that says:

 

if (!$_POST['submit'])

{show form}

else

{process mail form}

 

I tried using

if (!$_POST['login'])

on my login include instead of

if($_SERVER[REQUEST_METHOD]=='POST')

but that screwed with my include and caused an error a few lines down

[Daniel0]

[...] and caused an error a few lines down

 

When asking people for help when getting an error, it's often a good idea telling what the error is.

[jrobles]

sorry, I suck at life...

 

this is my error:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/vg001web02/68/11/3001168/web/dev/bsp/includes/login.inc.php on line 6

 

I am not sure why its looking for mysql data because the login form has not been submitted yet. This error happens right off the back regardless of what page I am on. Here are the first 15 lines of login.inc.php

 

<?
//if($_SERVER[REQUEST_METHOD]=='POST')
if (!$_POST['login'])
{
$sql="select * from products where id=$_POST[program]";
$program=mysql_fetch_assoc(mysql_query($sql));
//print_r($_POST);
//print_r($program);
$myAexxis=array(
	"q_system_key"=>$program[sitekey],
	"q_action"=>"CUST_AUTH",
	"q_cust_username"=>$_POST[username],
	"q_cust_password"=>$_POST[password]
	);
	$myCust=new CRM;

 

[Daniel0]

Right, that means your query failed, so you'll have to fix it. You could try echoing the query and see if it's generated correctly. You would also want to fix the SQL injection using mysql_real_escape_string, but that's unrelated to the error.

[jrobles]

Thanks a million for your help. I see the that the query is running but I have not submitted the form yet so it should not do anything until i hit the login button

[mrMarcus]

Thanks a million for your help. I see the that the query is running but I have not submitted the form yet so it should not do anything until i hit the login button

the ! before the $_POST['login'] would say otherwise.

 

that ! is like saying NOT:

 

<?php
if (!$_POST['login']) { //is saying, "if $_POST['login'] is NOT set, continue;
?>

 

lose the ! and the code will only execute IF $_POST['login'] has been set, and everything should be fine.

 

<?php
if (isset ($_POST['login'])) {
?>

 

EDIT:  see why posting your code and error(s) with a detailed description of what you are doing to get the error is crucial in getting help sooner than later.

[jrobles]

damn, that worked like a champ. those little details always get me. Thanks a million for your help guys