Why Does mysqli_stmt::bind_param() Throw 'Parameter ... expected to be a reference, value given'?

mysqli bind_param() Error: Passing Values Instead of References

The error "Parameter 3 to mysqli_stmt::bind_param() expected to be a reference, value given" arises when mysqli_stmt::bind_param() is passed an array of values instead of an array of references to those values.

PDO Code

In the PDO code provided:

$params = array(1,"2","3","4");

The $params array contains values rather than references to those values. To fix this, use & to create references:

$params = array(1,&"2", &"3", &"4");

OOP Code

Similarly, in the OOP code:

array_unshift($params, $param_type);

This line prepends the $param_type string to the beginning of the $params array, but it does not create references to the parameter values. To create references, use the & character:

array_unshift($params, $param_type);
foreach ($params as $key => $value) {
    $params[$key] =& $value;
}

Alternative Solution Using Spread Operator (PHP 5.6 )

In PHP versions 5.6 and above, you can use the spread operator (...) to pass the $params array as individual arguments without having to manually create references:

mysqli_stmt_bind_param($sql_stmt, $param_type, ...$params);

$insert_stmt->bind_param($param_type, ...$params);

By passing references or using the spread operator, you can ensure that mysqli_stmt::bind_param() receives an array of references to the parameter values, resolving the error.

The above is the detailed content of Why Does mysqli_stmt::bind_param() Throw 'Parameter ... expected to be a reference, value given'?. For more information, please follow other related articles on the PHP Chinese website!