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Bug #71476 Binding insert_id into prepared statement fails
Submitted: 2016-01-28 10:08 UTC Modified: 2017-05-21 04:22 UTC
Votes:1
Avg. Score:3.0 ± 0.0
Reproduced:0 of 0 (0.0%)
From: j dot nord at ntlworld dot com Assigned:
Status: No Feedback Package: MySQLi related
PHP Version: 5.5.31 OS: ubuntu4.14
Private report: No CVE-ID: None
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 [2016-01-28 10:08 UTC] 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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 [2017-05-10 10:18 UTC] fjanisze@php.net
-Status: Open +Status: Feedback
 [2017-05-10 10:18 UTC] 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
 [2017-05-21 04:22 UTC] php-bugs at lists dot php dot net 
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.
 
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