PHP and MySql

danhe

The Following is my PHP code

<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>

<?php require_once('Connections/testdatabase.php');
mysql_select_db("elementary", $testdatabase);
$search = mysql_query("select * from elementary", $testdatabase);
while($myarray=mysql_fetch_array($search))
{
echo $myarray["ID"];
echo $myarray["Location"]."<br>";
}
?>

<body>

</body>
</html>

The web server tells me that
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in f:\database\action.php on line 9
which is while($myarray=mysql_fetch_array($search))

I do not know why it is not working!!

Thanks for help!!

KevinMG

Originally posted by danhe
The Following is my PHP code

mysql_select_db("elementary", $testdatabase);
$search = mysql_query("select * from elementary", $testdatabase);

is both your database and the table you're trying to query named elementary??

also, side note...never hurts to use the special notation for tables inside queries...i.e.

tablename

Weedpacket

Replace two of the lines with these more informative variants:

mysql_select_db("elementary", $testdatabase)
    or die("Couldn't select database. MySQL reports: ".mysql_error());
$search = mysql_query("select * from elementary", $testdatabase)
    or die("Couldn't run query. MySQL reports: ".mysql_error());

planetsim

Error in your script weedpacket..

You cant select the same database twice.

Weedpacket

Er, I only see one database selection ... and one query.