MySql problem

Try rewriting your $sql = statment thus...


$sql = "Select * From Users where DomainLogin='".$RemoteUsersLogin."'";
   
and see what happens...

That isn't very easy to read...

$sql = [double quote]Select * From Users where DomainLogin=[single quote][doublequote].$RemoteUsersLogin.[double quote][single quote][double quote];
   

Could it be that you won't get a valid Userlogin from GetLoginName()?

RemoteUserLogin is IN FACT the correct username. I echoed it and it came up correctly, no spaces or extra characters or anything like that either. When I echoed the sql statement, it was correct and said "Select * from Users where DomainLogin = 'JHENDRIX'"

We suspect that it's a problem with permissions too. Any comments?

Ther.

how if you check the $IP. Did it value not 127.0.0.1 ? Make sure the $IP value like you need.

I modify DarinDB script below:

*****************************************
$IP = $REMOTE_ADDR;
$RemoteUsersLogin = GetLoginName($IP);
if ($RemoteUsersLogin) {
   $sql = "Select * From Users where DomainLogin='$RemoteUsersLogin'";
   $Result = mysql_query($sql,$MySqlConnection);
   if(empty($Result)){echo mysql_error();}  // <--- ADD THIS TO YOUR CODE
   else
   {
      if ($MyRow = mysql_fetch_array($Result)) {WORK!}
   }
*******************************************

Sorry DarinB ... my miss type your name above :)

Or display the $sql if error:

*****************************************
$IP = $REMOTE_ADDR;
$RemoteUsersLogin = GetLoginName($IP);
if ($RemoteUsersLogin) {
  $sql = "Select * From Users where DomainLogin='$RemoteUsersLogin'";
  $Result = mysql_query($sql,$MySqlConnection);
  if(empty($Result)){echo mysql_error()."<br>".$sql;}  // <--- ADD THIS TO YOUR CODE
  else
  {
     if ($MyRow = mysql_fetch_array($Result)) {WORK!}
  }
*******************************************

If you really got the correct username, the only solution is that it was not found in the database. Browse your database table and check if the username is in the database. For some reason something must have changed. If you need a quick code to browse your table try this:

==============================
  $sql = "select * from users";
  $result = mysql_query($sql, $MySqlConnection);
  show($result);

  function show($result)
  {
    if (!$result)
    {
        return;
    }

    if (0 == mysql_num_rows($result))
    {
        echo "nothing to show<br>\n";
        return;
    }

    $numfld = mysql_num_fields($result);
    echo "<table bgcolor=\"#CCFFFF\" border=\"1\">";
    for ($k = 0; $k < $numfld; $k++)
    {
        echo "<td>" . mysql_field_name($result, $k) . "</td>";
    }
    echo "</tr>\n";
    while ($items = mysql_fetch_row($result))
    {
      echo "<tr>";
      for ($k = 0; $k < $numfld; $k++)
      {
        echo "<td>" . $items[$k] . "&nbsp;</td>\n";
      }
      echo "</tr>\n";
    }
    echo "</table>\n";
  }
==============================

Another point could be your variable $MySqlConnection. Check if this is a valid database connection. Maybe your mysql_connect() failed in the first place for some reason and therefor you wouldn't get any results from the query. You should consider this, if you won't get a result from my code above.

Marian

Hi fajr n,

No problem on the mistype.

BTW, it is not necessary to test for an empty result since mysql_error() will be empty itself if no error occured, thus will not echo anything to the screen.


Heddesheimer,

It would not be a connection problem since PHP is actually sending the query to dB. If it was a connection problem, theravada would be getting back an invalid Link_Resource not an invalid Result Resource. Now, your suggestion may be valid if theravada IS getting an invalid Link Resource error instead.

darin

Linux is case-sensitive. So you should carefully check the Uppercase and lowercase of Tablenames, Fields and data field.

Adding the line:

echo mysql_error();

That showed me that the table name was users not Users, on NT I guess it didn't matter. Thanks!!

Ther.