Bug #71476 [Fbk->NoF]: Binding insert_id into prepared statement fails

From: Date: Sun, 21 May 2017 04:22:40 +0000
Subject: Bug #71476 [Fbk->NoF]: Binding insert_id into prepared statement fails
References: 1  Groups: php.bugs 
Request: Send a blank email to php-bugs+get-209200@lists.php.net to get a copy of this message
Edit report at https://bugs.php.net/bug.php?id=71476&edit=1

 ID:               71476
 Updated by:       php-bugs@lists.php.net
 Reported by:      j dot nord at ntlworld dot com
 Summary:          Binding insert_id into prepared statement fails
-Status:           Feedback
+Status:           No Feedback
 Type:             Bug
 Package:          MySQLi related
 Operating System: ubuntu4.14
 PHP Version:      5.5.31
 Private report:   N

 New Comment:

No feedback was provided. The bug is being suspended because
we assume that you are no longer experiencing the problem.
If this is not the case and you are able to provide the
information that was requested earlier, please do so and
change the status of the bug back to "Re-Opened". Thank you.


Previous Comments:
------------------------------------------------------------------------
[2017-05-10 10:18:57] fjanisze@php.net

I'm not able to reproduce your problem with the latest builds. I've been using a slightly
modified script:

.
$mysqli = new mysqli('localhost', 'root', '', 'test');

$mysqli->autocommit(false);
$stmt = $mysqli->prepare("INSERT INTO tab (id,name,age) VALUES(?,?,?)");
$plab = $mysqli->prepare("INSERT INTO labels (id) VALUES (?)");

$name = 'name';
$age = 10;

$stmt->bind_param("isi", $mysqli->insert_id, $name ,$age);
$stmt->execute();


$plab->bind_param("i", $db->insert_id);
$plab->execute();

if( $plab->error != '') {
    print 'error: '.$plab->error.PHP_EOL;
}
.

Can you verify again if on your environment this problem persist with recent releases? Thanks

------------------------------------------------------------------------
[2016-01-28 10:08:19] j dot nord at ntlworld dot com

Description:
------------
Using binding insert id into mysql prepared statement fails where a variable with the same value
works perfectly.

Test script:
---------------
http://pastebin.com/umgSYhrz

Expected result:
----------------
Code should echo:
Able to commit

Actual result:
--------------
Code echoes:
Error executing query: Column 'Barcode' cannot be null


------------------------------------------------------------------------



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Edit this bug report at https://bugs.php.net/bug.php?id=71476&edit=1


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