Question

相同的代码在另一个页面中正常工作，但现在它得到了以下错误。

mysqli_error（）只需要1个参数

当我回声它工作正常，但当我尝试插入它失败。不知道代码有什么问题。

<?php
session_start();
$id = $_SESSION["id"];
$cart_id = $_SESSION["cart_id"];
$cname = $_SESSION["cname"];
echo $cname;
echo $id;
echo $cart_id;
if (isset($_SESSION['cart_id'])) {
$con=mysqli_connect("localhost","root","","medical");

if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
error_reporting(E_ALL ^ E_NOTICE);
     $insert="INSERT INTO order(cart,cid,name)
VALUES($cart_id,$id,$cname)";

if (!mysqli_query($con,$insert)) 
    {
        die('Error: Failed Inserting' . mysqli_error());
    } 

echo "Success Fully Added ";
    }else{
        echo "somethimg wrong loser";
    }

?>

Answer 1

我甚至不想回答这个问题，因为我在上面的评论中为您提供了所有有用的链接，

这些是非常有用的链接

When to use single quotes, double quotes, and backticks in MySQL

How to get mysqli error in different environments?

Syntax error due to using a reserved word as a table or column name in MySQL

您的查询存在问题：

$insert="INSERT INTO order(cart,cid,name)
VALUES($cart_id,$id,$cname)";

您将这些变量视为整数，但$ cname是一个字符串，因此它应该是带引号的包装。

目前上面的代码将会出现

'字段列表'
中的未知列'WhateverValue of cname'

这应该是这样的

$con = mysqli_connect("localhost", "root", "", "medical");

    if (!$con) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    error_reporting(E_ALL ^ E_NOTICE);
    $insert = "INSERT INTO `order`(cart,cid,name) VALUES($cart_id,$id,'$cname')";


        $query = mysqli_query($con, $insert);
    if ($query) {

        echo "Success Fully Added ";
    }else{

        die('Error: Failed Inserting' . mysqli_error($con));
    }

编辑更好的方法是使用预准备语句

以这种方式，你不会像现在这样有简单的问题

$con = new mysqli("localhost", "root", "", "schedules");
    if ($con->connect_error) {
        die("Connection failed: " . $con->connect_error);
    } 


    $insert = "INSERT INTO `order`(cart,cid,name) VALUES(?,?,?)";

    $query = $con->prepare($insert);

    $query->bind_param("iis",$cart_id,$id,$cname);

    if($query->execute()){

         echo "Success Fully Added ";
    }else{

        die('Error: Failed Inserting' . $con->error);
    }

INSERT在相同代码在另一个文件中工作的地方不起作用

1 个答案: