ID: 46144
Comment by: Progman2002 at gmx dot de
Reported By: Progman2002 at gmx dot de
Status: Open
Bug Type: MySQLi related
Operating System: Linux
PHP Version: 5.2CVS-2009-01-25 (snap)
New Comment:
Sure I can call close() by myself (which I normally do), but it doesn't
solve the bug itself. The point is you get an error situation and don't
know why.
Previous Comments:
------------------------------------------------------------------------
[2009-05-18 02:17:56] felix9x at yahoo dot com
It's because the first $stmt object is destroyed by the second
assignment (which clears the last error message).
$sql = 'INSERT INTO
SomeTest(UserID, RechtID)
WHERE
(?,?)';
if (!$stmt = $db->prepare($sql)) {
die($db->error."-".$db->errno."-".$db->info);
}
Its equivalent to doing this: $stmt = false;
The destructor of the Mysqli_stmt class resets the Last error.
Its possible to call $stmt->close() explicitly. Probably best to use
this syntax:
$sql = 'INSERT INTO
t(i)
WHERE
(?)';
$stmt = $db->stmt_init();
if(!$stmt->prepare($sql) ){
die( $stmt->error );
}
------------------------------------------------------------------------
[2009-01-25 10:50:41] Progman2002 at gmx dot de
The bug is still not fixed. Maybe it has something to do with an
uncalled destructor since I use the same variable $stmt.
Actual result:
--------------
PHP-Version: 5.2.9-dev
MySQL-Server-Version: 50060
MySQL-Protocol: 10
-0-
Expected result:
----------------
PHP-Version: 5.2.9-dev
MySQL-Server-Version: 50060
MySQL-Protocol: 10
(Showing a MySQL error which says "Syntax error near WHERE (?,?)" or
says sth. like "unfinished prepare statement before")
------------------------------------------------------------------------
[2008-11-29 10:59:24] Progman2002 at gmx dot de
As the paste on the pasteboard is gone I'll add the code here.
-----------
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$db = @new MySQLi('localhost', '', '', 'test');
if (mysqli_connect_errno()) {
die('Konnte keine Verbindung zu Datenbank aufbauen, MySQL meldete:
'.mysqli_connect_error());
}
echo 'PHP-Version: '.PHP_VERSION."\n";
echo 'MySQL-Server-Version: '.$db->server_version."\n";
echo 'MySQL-Protocol: '.$db->protocol_version."\n";
$sql = 'CREATE TEMPORARY TABLE SomeTest(UserID INT NOT NULL, RechtID
INT NOT NULL)';
if (!$db->query($sql)) {
die($db->error);
}
$sql = 'DELETE FROM
SomeTest
WHERE
UserID = ?';
if (!$stmt = $db->prepare($sql)) {
die($db->error);
}
// note the missing $stmt->close() here
$sql = 'INSERT INTO
SomeTest(UserID, RechtID)
WHERE
(?,?)';
if (!$stmt = $db->prepare($sql)) {
die($db->error."-".$db->errno."-".$db->info);
}
echo "done";
?>
------------------------------------------------------------------------
[2008-09-21 13:49:28] Progman2002 at gmx dot de
Description:
------------
If you create a prepared statement with a DELETE query and tries to
create a second prepared statement with an INSERT query on the same
table without closing the first one the MySQLi::prepare() method failed,
but the fields $error and $errno (and all other related to them) aren't
filled with the error message. This is strange as the prepare() failed
but you dont know why.
The mysql error is shown if I save the second statement into another
variable (like if (!$stmt2 = $db->prepare($sql))) (maybe its related to
bug #44766)
Reproduce code:
---------------
Code is at http://nopaste.php-quake.net/51976
Expected result:
----------------
PHP-Version: 5.2.6-pl7-gentoo
MySQL-Server-Version: 50042
MySQL-Protocol: 10
{Showing a MySQL error which says "Syntax error near WHERE (?,?)" or
says sth. like "unfinished prepare statement before")
Actual result:
--------------
PHP-Version: 5.2.6-pl7-gentoo
MySQL-Server-Version: 50042
MySQL-Protocol: 10
-0-
(so the values are all empty strings or zero)
------------------------------------------------------------------------
--
Edit this bug report at http://bugs.php.net/?id=46144&edit=1
