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u0206787@nus.edu.sg

MySQL_Close error

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[code]
$connect = mysql_connect($host, $user, $password);
if (!$connect) {
   die('Not connected : ' . mysql_error());
}
$db=mysql_select_db($database, $connect);
if (!$db) {
   die ('Can\'t use $database: ' . mysql_error());
}
mysql_close($connect);
[/code]
Pls find what wrong with the code, I keep receive these errors.

These are the two error I received.
1. Undefined variable: connect

2. Warning: mysql_close(): supplied argument is not a valid MySQL-Link resource in c:\program files\easyphp1-8\www\test\login.php on line 86

Cheers.

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you should have to put
[code]$db=mysql_select_db($database, $connect);[/code]
you can just put:
[code]mysql_select_db($database, $connect);[/code]

and when using mysql_close you dont have to specify the link identifier. it can simply be:
[code]mysql_close();[/code]
and it will close the last open connection.

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[quote author=ProjectFear link=topic=109943.msg443614#msg443614 date=1159529620]
you should have to put
[code]$db=mysql_select_db($database, $connect);[/code]
you can just put:
[code]mysql_select_db($database, $connect);[/code]

and when using mysql_close you dont have to specify the link identifier. it can simply be:
[code]mysql_close();[/code]
and it will close the last open connection.
[/quote]

Thanks, however this time it showing another error. This time, it's showing:
"Warning: mysql_close(): no MySQL-Link resource supplied ".
I'm running on EasyPHP 1.8.0.1; with Apache 1.3.33; PHP 4.3.10; MySQL 4.1.9.

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[code]
<?php
if (isset($signin) && !empty($signin)) {

$connect = mysql_connect($host, $user, $password);
if (!$connect) {
  die('Not connected : ' . mysql_error());
}

$db=mysql_select_db($database, $connect);
if (!$db) {
  die ('Can\'t use $database: ' . mysql_error());
}

$result = mysql_query("SELECT * FROM Registration where studentid = '$studentid' and password='$pwd'")
or die(mysql_error());

mysql_close();
[/code]

Can anyone help me Pls?
"Warning: mysql_close(): no MySQL-Link resource supplied ".
I'm running on EasyPHP 1.8.0.1; with Apache 1.3.33; PHP 4.3.10; MySQL 4.1.9.

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[quote author=Gaoshan link=topic=109943.msg444106#msg444106 date=1159593097]
[code]mysql_close($connect)[/code]
[/quote]
when I use "mysql_close($connect);" it give me 2 errors as below.
1. Notice: Undefined variable:
2. Warning: mysql_close(): supplied argument is not a valid MySQL-Link resource in

however if I were to use "mysql_close();" it gave me an error as:
1. Warning: mysql_close(): no MySQL-Link resource supplied .

What wrong with this quote?

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[code]
if (isset($adminsignin) && !empty($adminsignin)) {
$connect = mysql_connect($host, $user, $password);
if (!$connect) {
  die('Not connected : ' . mysql_error());
}
$db=mysql_select_db($database, $connect);
if (!$db) {
  die ('Can\'t use $database: ' . mysql_error());
}
$result = mysql_query("SELECT * FROM Admin where AdminID = '$adminid' and Password='$pwd'")
or die(mysql_error());
if ($row = mysql_fetch_array($result))
{
}
else
{
}
}
mysql_close();
?>
[/code]

No i had checked, the nos of open blacket is equal to the nos of close bracket. Other than bracket, where you see the error could be? thanks again.

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This code (the one you last posted which now has the correct number of brackets) should work fine assuming you have the values for your various variables properly defined somewhere and assuming you put something in the currently empty if...else statement.

I tested it (cut and past and then changed variables and selects to match my setup) and it returned data no problem.

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