Question

我正在尝试获取表省的id以将其作为外键插入表城市但是我的代码中存在问题，我无法将其作为FK插入城市表中。这是代码。

<?php
 include 'connection.php';
 $result=mysqli_query ($conn,"set character_set_results='utf8'"); 
 $province_name=$_POST["province_name"];
 $city_name = $_POST["city_name"];
 $street = $_POST["street"];


    //$id=$_POST["id"];

   $result = mysqli_query($conn,"INSERT INTO province (province_name) VALUES ('$province_name')");
   $id=$_POST["province_id"];

   $id= mysql_insert_id($conn);
   mysql_free_result( $result );
    $result = mysqli_query($conn,"INSERT INTO city (pidfk,city_name, street) VALUES ('$id','$city_name','$street')");



    $res=array();
    //$res['check']=false;
   if(mysqli_multi_query($conn,$result))
     {
        $res['check']=true;

     }

header('Content-Type: application/json');
echo json_encode($res);
 ?>

Answer 1

您在使用mysql_insert_id时正在使用mysqli。你不能同时使用

将mysql_insert_id更改为mysqli_insert_id

我不能将一个表的主键作为外键发送给另一个表？

1 个答案: