A mysql_fetch_array error :(

[markman]

Hey all.
I am having some trouble with a mysql_fetch_array statement in a script (bellow) that should make news from a message board appear on the screen.

PHP Code:

<?php
include("nav.php");
?>
    <td width="610" valign="top" border="1">
      <table width="610" cellpading="0" cellspacing="0" align="left" border="0">
        <tr>
          <td width='610' height='9' bgcolor='#000000' valign='top'></td>
        </tr>
        <tr>
          <td width='610' height='13' background='images/index_06.gif' valign='top'></td>
        </tr>
        <tr>
          <td>
            <table width='610' border='0' rowspan='0' background='images/index_08.gif' >
<?php
include("connections/forum.php");
$get = mysql_query("SELECT * FROM `threads` WHERE `type` = 'announcement' ORDER BY id");
while ($array = mysql_fetch_array($get)) {
  $poster = $array['poster'];
  $get2 = mysql_query("SELECT * FROM users WHERE username='$poster'");
  while ($array2 = mysql_fetch_array($get2)) {
    echo "<tr>
                <td width='15%'><b><u>$array[poster]</u></b></a></td>
                <td width='84%' align='right'>$array[posttime]</td>
              </tr>
              <tr>
                <td align='center' valign='middle'>
                <img src='images/admin.jpg' alt='' border='2' width='80' height='80' />
              </td>
              <td width='84%' valign='top'><b><u>$array[title]</u></b><br /><br /><br /><br />\n
              $array[message]<br /><br />
              ________________________________________________________________
              </td>
            </tr>
";
  }
}
?>
          </table>
        </td>
       </tr>
       <tr>
         <td width='610' height='13' background='images/index_06.gif' valign='top' border='1'></td>
      </tr>
    </td>
  </table>
  </td>
  <td width="10" height="542" bgcolor="#000000" valign="top" border="1"></td>
  <td width="127" height="2" colspan="0" valign="top" background="images/index_04.gif" rowspan="16" align="top">
      <table cellpading="0" cellspacing="0" align="top" border="1">
        <tr>
          <td valign="top" rowspan="2" border="1">
          <?php
          if (!$_SESSION['userid']) {
          ?>
          <center><b>Log in</b></center>
          <form method="POST" action="login2.php">
          Username :<br />
          <input type="text" name="username" title="username" id="username" size="15" /><br />
          Password :<br />
          <input type="text" name="password" title="password" id="password" size="15" />
          <input type="submit" value="Log in" /><input type="reset" value="reset" />
          </form>
          <?php
          } else {
          ?>
          Welcome back : <br /><b><?php echo $_SESSION[username]; ?></b><br /><br />
          <img src="<?php echo $_SESSION[avatar]; ?>" alt="" border="2" /><br /><br />
          <a href="modifyprofile.php?id=<?php echo $_SESSION[userid]; ?>">
          Modify your profile</a>
          <?php
          }
          ?>
          </td>
        </tr>
      </table>
    </td>
  </tr>
</table>
</body>
</html>

Line 21 is where the error is, it says it isn't a valid mySql ressource
Line 21 is :
$get2 = mysql_query("SELECT * FROM users WHERE username='$poster'");

[doyle_loader]

I would try this:
$get2 = "SELECT * FROM users WHERE username='$poster'";

I don't use mysql_query(); for any of my select statements. But I am fairly new at this game.

Cheers,

Chris

[Kid Charming]

You have to use mysql_query() -- that's how the statement gets sent to the database. Without it, you've just got a string variable floating around. I do prefer, though, to keep my statements in a separate variable and send that to mysql_query() -- that way, it's easy to echo the query to see exactly what you're sending.

For some reason, your query is failing, so $get2 is being set to FALSE instead of to a resource. Try echoing your query to see what's being sent and adding 'or die(mysql_error())' to the end of your mysql_query() call to get MySQL's error message regarding the query.

[doyle_loader]

Oh Geez. I didn't even realize what I posted haha.

I always place SELECT * Blah Blah in a variable then use mysql_query();

Like this:

$sql = "SELECT CourseName, TeacherName, NotesTitle, FileName
		From Notes
		Where CourseName = '$CourseName'
		Order by NotesTitle Asc";

$result = mysql_query($sql);