Hello, I have a problem with this script when i will check it in the browser, the output is :
"Warning: mysql_query() expects parameter 1 to be string, resource given in C:\xampp\htdocs\doto\php\new_case.php on line 8
FOUT
SELECT * FROM disp_type ORDER BY t_naam ASC
Resource id #3"

<?php
 $host="localhost";
$username="TEST";
$password ="";
$dbnaam="DOTO";
 
 $db=mysql_connect($host, $username, $password) or die (mysql_error());
 mysql_select_db($dbnaam, $db) or die (mysql_error());

 $query_test = "SELECT * FROM disp_type ORDER BY t_naam ASC";
 if(!$result = mysql_query($db,$query_test))
 {
	 echo('FOUT<br>');
	 echo($query_test . '<br>');
	 echo($db);
	 exit;
 }
 while($record = mysql_fetch_array($result))
 {
	 echo "ID : {$record['t_id']}<br>Naam: {$record['t_naam']}<br>";
 }
 ?>

As you can see, echo($db) = Recource ID #3 ??
Can you please help me?

Recommended Answers

Your parms are reversed and the databse parm can be left out if you wish to use the last link open. Most pages only open one link anyway

The Syntax (from the manual) is:
resource mysql_query ( string $query [, resource $link_identifier ] )

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All 4 Replies

sorry line 11

SOLVED

Your parms are reversed and the databse parm can be left out if you wish to use the last link open. Most pages only open one link anyway

The Syntax (from the manual) is:
resource mysql_query ( string $query [, resource $link_identifier ] )
http://php.net/manual/en/function.mysql-query.php

You should write like this.

if(!$result = mysql_query($query_test,$db))

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