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mysqli – Php mysqi bind_param Number of variables doesn't match number of parameters in prepared statement

Posted by: admin April 23, 2020 Leave a comment

Questions:

This has to be a newbie mistake, but I’m not seeing it. Here is a snippet from my code:

$mysqli = mysqli_connect($dbCredentials['hostname'], 
    $dbCredentials['username'], $dbCredentials['password'], 
    $dbCredentials['database']);

if ($mysqli->connect_error) {
    throw new exception( 'Connect Error (' . $mysqli->connect_errno . ') '
    . $mysqli->connect_error);
}

$stmt = $mysqli->prepare("SELECT DISTINCT model FROM vehicle_types 
    WHERE year = ? AND make = '?' ORDER by model");

$stmt->bind_param('is', $year, $make);

$stmt->execute();

When I echo out the values for $year and $make, I am seeing values, but when I run this script, I get a null value, and the following warning appears in my log file:

PHP Warning:  mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement

In this case, year is in the database in type int(10), and I have tried passing a copy that had been cast as an int, and make is a varchar(20) with the utf8_unicode_ci encoding. Am I missing something?

How to&Answers:

Your prepared statement is wrong, it should be:

$stmt = $mysqli->prepare("SELECT DISTINCT model FROM vehicle_types WHERE year = ? AND make = ? ORDER by model");

The single quotes made that ? be the value not a marker. It will already be a string because you are casting as such with bind_param('is'