Skip to content
Advertisement

Return an array from mysqli stmt query

I’m trying to convert a site to use prepared mysql statements, however I’m having some trouble replacing my use of fetch_array().

The comments on php.net offer a solution which I would have thought should work but for some reason I’m getting strange undefined constant errors when calling a function via call_user_func_array() can someone suggest a better way of doing it?

Here’s the function I’m using at the moment. I’ve dumbed it down a bit for posting on here (it’s part of a larger class which extends mysqli and handles error catching etc) so sorry if I’ve made any mistakes copy and pasting it.

public static function stmt_bind_assoc(&$stmt, &$out) {
    $data = mysqli_stmt_result_metadata($stmt);
    $fields = array();
    $out = array();

    $fields[0] = $stmt;
    $count = 1;

    while($field = mysqli_fetch_field($data)) {
        $fields[$count] = &$out[$field->name];
        $count++;
    }
    call_user_func_array(mysqli_stmt_bind_result, $fields);
}
public function fetch_array($rawQuery) {
    $stmt = $this->prepare($rawQuery);
    $row = array();
    self::stmt_bind_assoc($stmt, $row);
    $result = $stmt->fetch();
    $stmt->free();
    return $result;
}

The errors I get are:

Notice: Use of undefined constant mysqli_stmt_bind_result – assumed ‘mysqli_stmt_bind_result’

Fatal error: Call to undefined method mysqli_stmt::free()

Advertisement

Answer

The first error comes from your call_user_func_array statement: you need quotes around the mysqli_stmt_bind_result, otherwise PHP assumes it’s a constant.

And there’s no free() function in mysqli_stmt, use $stmt->free_result() instead.

User contributions licensed under: CC BY-SA
1 People found this is helpful
Advertisement