mysql_fetch_array()

[ChrisDavis]

Hi, I'm having a nightmare, Why does this keep coming up, I'm trying to display the contents of a Table, but it keeps coming up with this:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/home/stcontractcom/public_html/view.php on line 20

It can be found at:
www.1stcontract.com

And here is my code:

<?

include('standard.php');

$dbh = mysql_connect($host, $user, $pass);
mysql_select_db($name);

$MasterTable = MasterTable;

//echo $MasterTable;

$stmt = "SELECT * FROM $MasterTable";
//echo $stmt;
$sth = mysql_query($stmt, $dbh);
//echo $sth;

//need to parse the variables from 2 behind, thats why it does not
//recognise the variable.

while($row = mysql_fetch_array($sth, MYSQL_ASSOC)) {

 echo $row["ID"];
 echo $row["Surname"];
 echo $row["Forename"];
 echo $row["LinktoName"];

}

// mysql_free_result ($sth);
mysql_close ($dbh);

?>

Any help will be gratefully appreciated.

Thanks

[saintp]

My first suggestion would be to quote 'MasterTable'; i.e.,

$MasterTable = "MasterTable";

There might be some funk there. Or you could just do

$stmt = "SELECT * FROM MasterTable";

That's all I see wrong with it -- and the barestring would not be flagged as an error, since PHP would interpret it as an (empty) constant.

[tomhath]

I hope you set $name somewhere in the include file.

$dbh = mysql_connect($host, $user, $pass) or die ("Connect failed: " . mysql_error()); 

mysql_select_db($name) or die ("Select failed: " . mysql_error()); 

$MasterTable = "MasterTable"; 

//echo $MasterTable; 

$stmt = "SELECT * FROM $MasterTable"; 

//echo $stmt; 
$sth = mysql_query($stmt, $dbh) or die ("Query failed: " . mysql_error());