Seleting from a row.

[JamieThompson]

Its been a while since I coded a project. I even forgot my account name for PHPbuilder 😛.
The problem is that I have a very large table with over 30 odd collums for a competiton Im starting up.
The thing is I want to search the whole row defined by CID

$comp = $_GET['comp'];
require_once ('connect.php');
$query = "SELECT * FROM `psrelay` WHERE `CID` = $comp";

However I will be soon implementing a members section where userid will be stored in the session. I want to search the whole record where CID = $comp for the user ID. What would the easiest way to achive this, or will I have to make mulitiple OR statments defining my collums?
Any help apriciated,
Thanks - Jamie Thompson

[iNFERiON]

If you store the userid in a session, and want to know wheter the userid exists in a row you should be doing something like this.

if (!session_id()) {
  session_start();
}

$comp = $_GET['comp'];
$userid = $_SESSION['userid'];
require_once ('connect.php');
$query = "SELECT * FROM `psrelay` WHERE `CID` = $comp";
$arr = mysql_fetch_array($query);
foreach ($arr as $check) {
  if ($check==$userid) {
    echo "User ID Found !";
  } else {
    echo "Nothing here";
  }
}

Hope that's what you're looking for.

[JamieThompson]

Thats more or less what im trying to achive,
I recive a fair few errors though. I have yet to use mysql_fetch_array, I have done a bit of reading and I tried to implement it myself.
The origial code gave me this error.

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\Jamie Thompson\My Documents\www\Digitally-United\memberincontest.php on line 6

Warning: Invalid argument supplied for foreach() in C:\Documents and Settings\Jamie Thompson\My Documents\www\Digitally-United\memberincontest.php on line 7

I changed the code abit so I could use GET for the user id,

<?php
$comp = $_GET['comp'];
$mem = $_GET['mem'];
require_once ('connect.php');
$query = "SELECT * FROM `psrelay` WHERE `CID` = $comp"; 
$result = mysql_query ($query);
if ($result)
{
while ($arr = mysql_fetch_array ($query, MYSQL_ASSOC))
{ 
foreach ($arr as $check) { 
  if ($check==$userid) { 
    echo " You are in the competition: ID $row[CID], Name $row[Name]";
  } else { 
    echo "You are not in competition: ID $row[CID], Name $row[Name] "; 
  } 
}
};
};
?>

Now I get

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\Jamie Thompson\My Documents\www\Digitally-United\memberincontest.php on line 9

Could someone give me some adivce on wher I am going wrong as I have yet to come accross this error before.
Thanks.

[mjax]

This error means that either you have an error in your SQL code or you are not connected to the database. Either way, the specific error is easy to obtain. Just use the next statement, directly after the mysql_query() statement:

if(!$result) die('ERROR: '.mysql_error());

[TheDefender]

Right under your query echo it to see what it's value is and to make sure it's what you expect it to be.

echo $query;