error part is $queryReference->bind_param(“s”,$_POST[‘reference_number’])
$sqlReference = "SELECT count(id) as reference_cnt FROM payments WHERE reference_number = ?";
$queryReference = $db->prepare($sqlReference);
$resultReference = $queryReference->get_result();
$rowReference = $resultReference->fetch_assoc();
array_push($errors,"This Reference Number already exists");
The error is because the prepare() call failed and you don’t have any error handling to tell you if and why a database statement failed.
This code is attempting to do the same thing as in one of your previous threads - Trying to check if block_lot_no is already existing using prepared statements I recommend that you reread that thread and do the things mentioned in the replies, concerning just attempting to INSERT the data and detecting in the error handling if a duplicate index error occurred. The suggested usage of exceptions for database statement errors will also tell you why the prepare call is failing in this code.
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