MySQL problens

[ddrummond]

I construct this user authenticate:

if (!($result = mysql_query("$db", "select * from user_profile where user_id ='$form_user_id'")))
{
DisplayErrMsg(sprintf("internal error %d:%s\n", mysql_errno(), mysql_error()));
return 0;
}

if (($row = mysql_fetch_array($result)) && ($form_password == $row ["password"] && $form_password != ""))
	return 1;
else
	return 0;

and in the brower display this:

Warning: Supplied argument is not a valid MySQL-Link resource in E:\Projectos\secla\backoffice\functions.php on line 26

internal error 0:

Help me please !!!! i don't know how argument is not valid.

[mithril]

Daniel,

mysql_query() accepts an sql query string (and optionally a connection resource identifier). It's mysql_db_query() which accepts the database name, sql query, and the optional connection resource identifier. However, as of PHP 4.0.6 mysql_select_db() is considered deprecated. The proper way is:

  mysql_select_db($db); // note the absense of quotes
  $result = mysql_query("select * from user_profile where user_id ='$form_user_id'");

HTH.

geoff

http://www.php.net/manual/en/function.mysql-db-query.php
http://www.php.net/manual/en/function.mysql-query.php

[ddrummond]

Tanks, very much