php - mysqli_stmt::bind_param() 的参数 2 预期为引用，给出值

Question

$type = array('i','i');
$param = array("1","1");
$stmt = $mysqli->prepare("SELECT par1, par2 FROM table WHERE par3 = ? AND par4 = ?");
$refs = array();
foreach($param as $key => $value) {
        $refs[$key] = &$param[$key];
}
$result_params = array_merge($type,$refs);
call_user_func_array(array($stmt, 'bind_param'), $result_params);
$stmt->execute();

---

var_dump($result_params):

array(4) {
  [0]=>
  string(1) "i"
  [1]=>
  string(1) "i"
  [2]=>
  &string(1) "1"
  [3]=>
  &string(1) "1"
}

---

当我们使用代码时，我们会收到错误:

Parameter 2 to mysqli_stmt::bind_param() expected to be a reference, value given in ...

为什么我们会收到此错误以及如何解决该问题？

Answer 1

问题已解决。

解决问题:

相反，$type = array('i','i');需要使用$type = array('ii');