ID: 46144
Updated by: [email protected]
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:
I don't know what to do with this report because in a way it is a user
error. The problem can easily be avoided by calling the destructor of
the mysql_statement class before assigning false to it. A simple
$stmt->close() before the second $stmt = $db->prepare() will fix it.
What happens is:
$stmt = ... <valid_sql>
$stmt = $mysqli->prepare('wrong_sql');
<switch into php />
<function_enter>
C API: mysql_stmt_init()
C API: mysql_stmt_prepare() -> error
copy error, because mysql_stmt_close() will clean it
C API: mysql_stmt_close()
RETURN false
</function_enter>
<evaluate assignment to $stmt>
$stmt is a Prepared Statement for <valid_sql>
destruct $stmt, because user did not clean it up!
C API: mysql_stmt_close()
!!! previously saved error message gets cleaned !!!
</evaluate assignment to $stmt>
<assign return value to $stmt />
<give control back to PHP script />
var_dump($mysqli->error) -> no error message
One way to fix on the C level would be to blow up the MY_MYSQL struct
and copy the error message into some safe place. But I wonder how we
would know when to return the message copied into a safe place and when
not... currently I can't think of a way how we would know.
It may be possible to hack something with mysqlnd and make mysqlnd
behave different to libmysql (not clean/preserve error message in
mysqlnd' stmt dtor) but that would be just the wrong place and really
hackish....
Maybe we should set the status to "Won't fix".
Previous Comments:
------------------------------------------------------------------------
[2009-05-18 14:43:14] Progman2002 at gmx dot de
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.
------------------------------------------------------------------------
[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
