Passing through UserID

[jarv]

here is my code:

$query2 = "SELECT * FROM members_copy WHERE RSUSER = ".$_SESSION["rsUser"]."";
$result2 = mysql_query($query2);
$UserID = $result2['USERID'];
echo $UserID;

 

 

When I echo out UserID it doesn't show a number?!

 

Please help?!

[Pikachu2000]

$result2 doesn't hold a value, it holds a query result resource. You need to get the values from that with mysql_fetch_assoc or similar.

[jarv]

thanks, now I get: <b>Warning</b>:  mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in <b>D:\mypubspace.com\wwwroot\main.php

[jarv]

please help

[Pikachu2000]

Post your newly revised code.

[jarv]

$data = "SELECT * FROM members_copy WHERE RSUSER = ".$_SESSION["rsUser"]."";

$info = mysql_fetch_array( $data ); 
echo $info['USERID'];

[PFMaBiSmAd]

The following link is a basic php/mysql tutorial that shows how to execute a select query and fetch the results from the query - http://w3schools.com/php/php_mysql_select.asp

[jarv]

Please help?!

is this correct?

$data = "SELECT * FROM members_copy WHERE RSUSER = ".$_SESSION["rsUser"];

$info = mysql_fetch_array($data); 

[jarv]

I basically want to get the USERID of the person logged in

[jarv]

here is my login functions:

function checkPass($login, $password) {
/*
Password checking function:
This is a simple function that takes the $login name and
$password that a user submits in a form and checks that a
row exists in the database where:

the value of the 'login' column is the same as the value in $login
and
the value of the 'password' column is the same as the value in $password

If exactly one row is returned, then that row of data is returned.
If no row is found, the function returns 'false'.
*/
global $link;

$query="SELECT rsUser, rsPass FROM members WHERE rsUser='$login' and rsPass='$password'";
$result=mysql_query($query, $link)
	or die("checkPass fatal error: ".mysql_error());

// Check exactly one row is found:
if(mysql_num_rows($result)==1) {
	$row=mysql_fetch_array($result);
	return $row;
}
//Bad Login:
return false;
} // end func checkPass($login, $password)


/******************************************************\
* Function Name : cleanMemberSession($login, $pass)
*
* Task : populate a session variable
*
* Arguments : string $login, string $pass
			taken from users table in db.
*
* Returns : none
*
\******************************************************/
function cleanMemberSession($login, $password) {
/*
Member session initialization function:
This function initializes 3 session variables:
  $login, $password and $loggedIn.

$login and $password are used on member pages (where you
could allow the user to change their password for example).

$loggedIn is a simple boolean variable which indicates
whether or not the user is currently logged in.
*/

$_SESSION["rsUser"]=$login;
$_SESSION["rsPass"]=$password;
$_SESSION["loggedIn"]=true;
} // end func cleanMemberSession($login, $pass)

[jarv]

Please help?

[BlueSkyIS]

Please help?!

is this correct?

$data = "SELECT * FROM members_copy WHERE RSUSER = ".$_SESSION["rsUser"];

$info = mysql_fetch_array($data); 

 

what happens when you TRY IT?

[devilinc]

i think the session variable stores alphanumeric data...

 

CHANGE:::

$data = "SELECT * FROM members_copy WHERE RSUSER = ".$_SESSION["rsUser"];

$info = mysql_fetch_array($data); 

 

DO THIS:::

//note that i have added quites befire and after your session variable...
$data = "SELECT * FROM members_copy WHERE RSUSER = ' ".$_SESSION["rsUser"]." ' ";
$info = mysql_fetch_array($data);
//say if user_id is the user id column in your table then
echo $info['usr_id'];