Question

    <?php
    session_start();
    $user = $_SESSION['login'];
    $mysql_connect = mysql_connect("localhost", "root", "");
    $mysql_select_db = mysql_select_db("site");
    $sql = "SELECT * FROM `msg_inbox` WHERE `to` = '$user'";
    $result = mysql_query($sql) or die(mysql_error());
    while ($row = mysql_fetch_array($result)) {    
            $mysql_connect = mysql_connect("localhost", "root", "");
            $mysql_select_db = mysql_select_db("site");
            $query = ("UPDATE msg_inbox  SET unread = 0 WHERE id= ".$row['id']);
            $result = mysql_query($query);
    }

And I receive error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in on line 8
Can you help me?

Answer 1

You have reassigned $result in your loop. After the first iteration, the variable $result will hold a boolean indicating the success or failure of your UPDATE statement.

Change:

$result = mysql_query($query);

To:

$result2 = mysql_query($query);

Or just:

mysql_query($query);

And be careful of SQL injection holes.

EDIT actually, your whole code could and should be shortened to:

<?php

  session_start();
  $user = $_SESSION['login'];
  // A blank password for root? Really?
  $mysql_connect = mysql_connect("localhost", "root", "");
  $mysql_select_db = mysql_select_db("site");
  $sql = "
    UPDATE `msg_inbox`
    SET `unread` = 0
    WHERE `to` = '".mysql_real_escape_string($user)."'
  ";
  // Don't show the result of mysql_error() in a production environment!
  $result = mysql_query($sql) or die(mysql_error());