supplied argument is not a valid MySQL

[Bowers007]

Here is the code:

<?php

$con = mysql_connect("localhost","user","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

$result1 = mysql_query("SELECT employee.LastName, employee.FirstName, annual.Emp_Id,  

annual.harassment, annual.bbp, annual.loto, annual.hazcomm, annual.ppe, annual.evacuation, annual.drug  

FROM employee, annual 

Where employee.Emp_Id=ms.Emp_Id 
AND employee.Emp_Id<>'60' 
AND employee.Emp_Id<>'68' 
AND employee.Emp_Id<>'8' 
AND employee.Emp_Id<>'67' 
AND employee.Emp_Id<>'80' 
AND employee.Emp_Id<>'79' 
AND employee.Emp_Id<>'75' 
AND employee.Emp_Id<>'56' 
AND employee.Emp_Id<>'76' 
AND employee.Emp_Id<>'5'
AND employee.Emp_Id<>'87' 
AND employee.Emp_Id<>'83' 
AND employee.Emp_Id<>'86'
AND employee.Emp_Id<>'84'
AND employee.Emp_Id<>'92'
AND employee.Emp_Id<>'81'
AND employee.Emp_Id<>'72'

ORDER BY employee.LastName
");

echo "<table border='1'>
<tr>
<th>Employee ID</th>
<th>Last Name</th>
<th>First Name</th>
<th>Harassment</th>
<th>BBP</th>
<th>LOTO</th>
<th>Haz Comm</th>
<th>PPE</th>
<th>Evacuation</th>
<th>TN Drug Free</th>
</tr>";

(LINE 89) while($row = mysql_fetch_array($result1))
  {
  echo "<tr>";
  echo "<td>" . $row['Emp_Id'] . "</td>";
  echo "<td>" . $row['LastName'] . "</td>";
  echo "<td>" . $row['FirstName'] . "</td>";
  echo "<td>" . $row['harassment'] . "</td>";
  echo "<td>" . $row['bbp'] . "</td>";
  echo "<td>" . $row['loto'] . "</td>";
  echo "<td>" . $row['hazcomm'] . "</td>";
  echo "<td>" . $row['ppe'] . "</td>";
  echo "<td>" . $row['evacuation'] . "</td>";
  echo "<td>" . $row['drug'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);

?>

Here is the error I get:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/Training/SearchResultsAnnual.php on line 89

Thanks for any help. I have several .php files that have the same error, this was working perfect but now errors. One of the files work perfect (basically the same code)

[NogDog]

That error usually indicates that MySQL did not like something about the query and rejected it, in which case mysql_query() returns false instead of a query result resource. Check the value of the variable returned by the mysql_query() call ($result1 in this case), and if it is false, get some debug info by either logging or ouputting the result of mysql_error().

[bradgrafelman]

First thing you should do when faced with a MySQL error is get all of the error information. Common suggestion you'll see (this same type of question gets asked probably a couple of times per week, by the way) is to write your code like so:

$query = 'SELECT col1 FROM foo...'; // store SQL query in a variable
$exec = mysql_query($query) or die('MySQL Error: ' . mysql_error() . "<br>Query: $query");

The "or die()" clause will let you see a) the error message MySQL returned, and b) the exact query string that was sent.

If you can't spot the error after getting that extra debugging information, copy and paste the info for us to see.

EDIT: NogDog beat me to it... :p