Bug #78932 [Opn->Ver]: Cannot fetch mysqli_prepare error if $statement variable reused

Edit report at https://bugs.php.net/bug.php?id=78932&edit=1 ID: 78932 Updated by: nikic@php.net Reported by: craig at craigfrancis dot co dot uk -Summary: mysqli_prepare oddity +Summary: Cannot fetch mysqli_prepare error if $statement variable reused -Status: Open +Status: Verified Type: Bug Package: MySQLi related -PHP Version: 7.4.0 +PHP Version: 7.3.12 Block user comment: N Private report: N New Comment: Also reproduces on older PHP versions. Previous Comments: ------------------------------------------------------------------------ [2019-12-08 21:27:49] craig at craigfrancis dot co dot uk Description: ------------ When running the following test script, the error is not correctly shown. But if you un-commented the $statement = false line, then it works. It's as though the $statement variable is not being properly replaced. Test script: --------------- <?php $link = mysqli_connect('localhost', 'username', 'password', 'database'); $statement = mysqli_prepare($link, 'SELECT 1'); // $statement = false; $statement = mysqli_prepare($link, 'SELECT 1 FROM this_table_does_not_exist'); if (!$statement) { exit(mysqli_errno($link) . ': ' . mysqli_error($link)); } ?> Expected result: ---------------- 1146: Table 'database.this_table_does_not_exist' doesn't exist Actual result: -------------- 0: ------------------------------------------------------------------------ -- Edit this bug report at https://bugs.php.net/bug.php?id=78932&edit=1