Check syntax for mysql_query() function. You don't need to pass the
database name. See here:
http://www.php.net/manual/en/function.mysql-query.php
You're query method then becomes:
function query($query) {
$connection = mysql_connect($this->hostname, $this->user,
$this->pass) or die ("Cannot connect to database");
mysql_select_db ($this->db, $connection); // You forgot this step
// Also change the following line, you don't need
// to pass the database name
$ret = mysql _query($query, $connection) or die ("Error
in query: $query");
return $ret;
}
Robert Zwink
http://www.zwink.net
-----Original Message-----
From: Navid Yar [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, March 06, 2002 4:59 PM
To: 'Robert V. Zwink'
Subject: RE: [PHP] Database Error
I just tried that Robert. It still gives me the same error that it did
before. There is no error after the mysql_connect. It definitely is
connecting. If it didn't connect then it wouldn't pass the query to the
database in the first place and spit out an error on the SQL syntax. I
checked the syntax on the MySQL command line and it worked just fine. So
it's not the syntax either. I don't know what it could be :(
-----Original Message-----
From: Robert V. Zwink [mailto:[EMAIL PROTECTED]]
Sent: Thursday, March 07, 2002 3:42 PM
To: Navid Yar
Subject: RE: [PHP] Database Error
Could you try changing your query() method to
function query($query) {
$connection = mysql_connect($this->hostname, $this->user,
$this->pass) or die ("Cannot connect to database");
echo mysql_error(); // put this here
$ret = mysql _query($this->db, $query, $connection) or die ("Error
in query: $query");
return $ret;
}
I think echoing out the mysql_error() will help pinpoint the problem.
Seems
like it is unable to connect?
Robert Zwink
http://www.zwink.net/daid.php
-----Original Message-----
From: Navid Yar [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, March 06, 2002 4:42 PM
To: [EMAIL PROTECTED]
Subject: RE: [PHP] Database Error
Yes, Jason. I posted it earlier, but here it is again.
// Error I'm getting
Error in query: SELECT label FROM menu WHERE id = 3
Warning: Supplied argument is not a valid MySQL result resource in
e:\localhost\menu\menu.class.php on line 47
// Object being called from get.php
<?php
require("menu.class.php");
$obj = new Menu();
echo $obj->get_label(1);
?>
// Class in the file menu.class.php
class Menu {
var $hostname;
var $user;
var $pass;
var $db;
var $table;
function Menu() {
$this->set_database_parameters("localhost", "username",
"password", "apps", "menu");
}
function set_database_parameters($hostname, $user, $password, $db,
$table) {
$this->hostname = $hostname;
$this->user = $user;
$this->password = $password;
$this->db = $db;
$this->table = $table;
}
function query($query) {
$connection = mysql_connect($this->hostname, $this->user,
$this->pass) or die ("Cannot connect to database");
$ret = mysql _query($this->db, $query, $connection) or die ("Error
in query: $query");
return $ret;
}
function get_label($id) {
$query = "SELECT label FROM $this->table WHERE id = $id";
$result = $this->query($query);
$row = mysql_fetch_row($result);
return $row[0];
}
}
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