> [new code]
> if (!mysql_connect($a, $b, $c)) return;
>
> if (!mysql_select_db($dbname)) return;
>
> $res = mysql_query("SELECT * FROM manual;", $link);
> [/new code]
Isn't going to work because $link is not defined and definitely not a
resource like mysql_query expects.
> OR, optionally, to surpress the warnings:
>
> [new code]
> if (!mysql_connect($a, $b, $c)) return;
>
> $link = null;
>
> if (!mysql_select_db($dbname)) return;
>
> $res = mysql_query("SELECT * FROM manual;", $link);
> [/new code]
Isn't going to work because mysql_query needs a resource to connect to.
You've defined $link as null.
$ cat test.php
<?php
$user = 'my_db_user';
$pass = 'my_pass';
$host = 'localhost';
$db = 'my_db';
error_reporting(E_ALL);
ini_set('display_errors', true);
if (!mysql_connect($host, $user, $pass)) {
die("unable to connect");
}
if (!mysql_select_db($db)) {
die("unable to choose db");
}
$link = null;
$res = mysql_query('select version() as version', $link);
while ($row = mysql_fetch_assoc($res)) {
print_r($row);
}
$ php test.php
Warning: mysql_query(): supplied argument is not a valid MySQL-Link
resource in /path/to/test.php on line 19
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL
result resource in /path/to/test.php on line 20
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