mysql_fetch_array() expects parameter 1 to be resource

[timstring]

I'm not sure I care, because the script works as is should. But, I am a perfectionist, which one has to be when programming dumb computers.

Here's the code:

<?php


$host="localhost";
$user="root";
$password="g1G3m#1989";
$Dbname = "DispatchReports";


$cxn = mysql_connect($host,$user,$password)
or die (mysql_error() . "couldn’t connect to server");


mysql_select_db($Dbname, $cxn)
or die(mysql_error() . "couldn't select database");


	$result = mysql_query("SELECT * FROM Techs", $cxn);
	$num_rows = mysql_num_rows($result);


for ($TechCount = 1; $TechCount <= $num_rows; $TechCount++)
	{
		$Quest = "SELECT * FROM `DispatchReports`.`Techs` WHERE IDNum = '$TechCount'";
		$result = mysql_query($Quest, $cxn)
			or die ('Santa Claus is watching you' . mysql_error());

		while ($row = mysql_fetch_array($result))
			{
				$FirstName = $row['FirstName'];
				$LastName = $row['LastName'];
				$TechNum = $row['TechNum'];
				$CellNum = $row['CellNum'];






				$AreaCode = substr($CellNum, 0, 3);
				$AreaCode = '(' . $AreaCode . ') ';

				$Exchange = substr($CellNum, 3, 3);


				$PhNum = substr($CellNum, 6);

				$CellNumber = $AreaCode . ' ' . $Exchange . '-' . $PhNum;

				echo $FirstName . ' ' . $LastName . ' ' . $CellNumber . '<br />';




				$result = mysql_query("UPDATE `DispatchReports`.`Techs` SET `CellNum` = '$CellNumber' WHERE `IDNum` = $TechCount")
					or die ("cannot update   ". $TechNum);

			}
}

The error is:

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /Applications/XAMPP/xamppfiles/htdocs/FixCell.php on line 40
Jerry Elliott [phone]

Line 40 is:

while ($row = mysql_fetch_array($result))

This error is repeated for every iteration of $TechCount. I am flummoxed because I copied most of the lines of code from a page that worked. I have a similar script, and I'm getting the same error in the same line, but again, the script produced the results I am after.

[Weedpacket]

Have you noticed that you're changing the value of [font=monospace]$result[/font] inside the loop?

[sneakyimp]

Yes you need a different variable name for your UPDATE query inside your loop. As your code currently operates, it first sets $result to be the results of a SELECT query but then the last line of your loop redefines it to be the result of an UPDATE query which is never a mysql resource but rather a boolean (TRUE or FALSE) value. Also, you should stop using the mysql functions because they have been deprecated. You should instead be using mysqli or PDO functions instead.