php - 尝试获取插入值后得到的返回 ID，但 mysqli_insert_id() 返回 null

Question

<?php 

 $value = json_decode(file_get_contents('php://input'));
 $mysql_pekare= new mysqli ("serv", "user","pass", "db");

 if(!empty($value)) {
 $stmt = $mysql_pekare->prepare("INSERT INTO users (`username`, `password`)  VALUES(?,?)"); 

 $stmt->bind_param("ss", $value->username, $value->password);
 $stmt->execute();

     if(!empty($stmt)) {

        $contacts = array(); 
        $id = mysqli_insert_id(); 
        $contact = array("objectId" =>  ($id));
        array_push($contacts, $contact);
        echo json_encode(array('results' => $contacts), JSON_PRETTY_PRINT);
     }

 $stmt->close();
 $mysql_pekare->close();

 }
 ?>

现在，当我从我的前端插入我的信息时，值(用户名 + 密码)被添加到 MySQL 中并且具有唯一 ID，但我没有得到返回的 ID，目前我是这样得到的:

   {
  "results": [
      {
  "objectId": null
      }
     ]
   }

Answer 1

您的 $id 很可能为 null 或不确定的错误...

//This is for procedural and needs a link
$id = mysqli_insert_id(); 
//should be
$id = mysqli_insert_id($mysql_pekare); 
//$mysql_pekare I am assuming is your connection...

//however you are using oo (or I think) so should be
$stmt->insert_id