Question

我在另一个表中插入数据后尝试生成singer id，当我使用准备语句时，问题就出现了：

警告：mysqli_insert_id（）期望参数1为mysqli，object＆gt;在/Applications/XAMPP/xamppfiles/htdocs/admin/singer.php中给出＆gt;第22行警告：mysqli :: prepare（）[mysqli.prepare]：无法获取＆gt;在/Applications/XAMPP/xamppfiles/htdocs/admin/singer.php上的mysqli＆gt;第24行

<?php
include('../db_inc.php');    
$singer_name =$_POST['singer_name'];
$singer_gender=$_POST['singer_gender'];
$singer_des=$_POST['singer_description'];
$singer_genre=$_POST['genre_list'];

        if($stmt=$connection->prepare("INSERT INTO singers(singer_name,singer_gender,singer_description) VALUES (?,?,?)")){
            $stmt->bind_param('sss',$singer_name,$singer_gender,$singer_des);
            $result1=$stmt -> execute();
            $singer_id=mysqli_insert_id($stmt);
            $stmt->close();
            }

  if($stmt2=$connection->prepare("INSERT INTO genre_singer(f_singer_id,f_genre_id) VALUES (?,?)")){
      $stmt2->bind_param('fs',$singer_id,$singer_genre);
            $result2=$stmt2 -> execute();
            $stmt2->close();
            $connection->close();
      }
if($result1 & $result2){
    echo "insert successfully"; 
    };
?>

Answer 1

变化：

$singer_id=mysqli_insert_id($stmt);

为：

$singer_id=mysqli_insert_id($connection);

Answer 2

您没有正确使用mysqli_insert_id。它应该是：

mysqli_insert_id($connection);

或者更好：

$connection->insert_id

请参阅文档：http://php.net/mysqli_insert_id

使用“准备声明”后未生成“mysqli_insert_id”

2 个答案: