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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
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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From: j dot nord at ntlworld dot com
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 [2016-01-28 10:08 UTC] j dot nord at ntlworld dot com
Using binding insert id into mysql prepared statement fails where a variable with the same value works perfectly.

Test script:

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]
-Status: Open +Status: Feedback
 [2017-05-10 10:18 UTC]
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');

$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);

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

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.
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change the status of the bug back to "Re-Opened". Thank you.
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