PHP mysqli 准备语句不工作

Question

我正在使用 mysqli prepare 语句来查询具有多个约束的数据库。我已经在我的测试文件中运行了代码，它工作得很好。但是，当我将代码移至我的实时文件时，它会抛出以下错误:

PHP Warning: mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement in C:\wamp\www\firecom\firecom.php on line 80

PHP Notice: Undefined variable: results in C:\wamp\www\firecom\firecom.php on line 89

两个参数都已正确设置，但有些东西将其丢弃。

代码:

$query = $mysqli->prepare("SELECT * FROM calls WHERE wcccanumber = ? && county = ?");
$query->bind_param("ss", $wcccanumber, $county);
$query->execute();

$meta = $query->result_metadata();

while ($field = $meta->fetch_field()) {
    $parameters[] = &$row[$field->name];
}

call_user_func_array(array($query, 'bind_result'), $parameters);

while ($query->fetch()) {
    foreach($row as $key => $val) {
        $x[$key] = $val;
    }
    $results[] = $x;
}

print_r($results['0']);

$查询变量转储:

object(mysqli_stmt)#27 (10) { ["affected_rows"]=> int(-1) ["insert_id"]=> int(0) ["num_rows"]=> int(0) ["param_count"]=> int(2) ["field_count"]=> int(13) ["errno"]=> int(0) ["error"]=> string(0) "" ["error_list"]=> array(0) { } ["sqlstate"]=> string(5) "00000" ["id"]=> int(1) }

Answer 1

为什么要用mysqli折磨自己？
在 PDO您将不需要这些可怕的代码，只需一行即可获得结果

$query = $pdo->prepare("SELECT * FROM calls WHERE wcccanumber = ? && county = ?");
$query->execute(array($wcccanumber, $county));
$results = $query->fetchAll();
print_r($results[0]);