查询在 PHPMyAdmin 中工作但在 PHP 中不工作

Posted 2023-02-25

【中文标题】查询在 PHPMyAdmin 中工作但在 PHP 中不工作

【英文标题】：Query working in PHPMyAdmin but not in PHP

【发布时间】：2015-05-17 12:56:18

【问题描述】：

我是 php 新手，已经学习了一个在线教程，到目前为止我一直工作正常，但现在我的数据库没有返回查询，但是当我去 PHPmyadmin 那里时，我可以让查询正常工作。

下面是代码

    <?php 
ob_start();
//Delete Item question to admin and delete product

include"../storescripts/connect_to_mysql.php";

if (isset($_GET['deleteid'])) 
    echo 'Do you really want to delete the item with ID '.$_GET['deleteid'].'?<a href="inventory_list.php?yesdelete='.$_GET['deleteid'].'">Yes</a>|<a href="inventory_list.php">No</a>';
    exit();
     

if(isset($_GET['yesdelete']))
    // Delete the actual product and delete picture also
    //delete from database
    //$id_to_delete = $_GET['yesdelete'];
    //echo  $id_to_delete;

     $sql =mysqli_query( "DELETE * FROM `products` WHERE `id`=2  LIMIT1 ");

    //mysql_query("DELETE * FROM `products` WHERE `id`='$id_to_delete'LIMIT1") or (mysql_error());

    //mysqli_query("DELETE * FROM products WHERE id=`$id_to_delete`LIMIT1");// or (mysql_error());

    //Unlink file from server
    $pictodelete=("../inventory_images/$id_to_delete");
    //echo $pictodelete;
    if(file_exists($pictodelete))
        unlink($pictodelete);
        

        header("location:inventory_list.php");
        exit();

       


?>

非常感谢您的帮助，我的服务器读取 PHP Extension :mysqli 。

【问题讨论】：

LIMIT1 应该是LIMIT 1，带有一个空格...

危险：你容易受到SQL injection attacks的影响，你需要自己去defend。 危险：此代码是vulnerable to XSS。用户输入在插入 html 文档之前需要转义！

【参考方案1】：

我不知道 connect_to_mysql.php 里面是什么，但起初有一个 连接到数据库的过程 我假设你做得正确，它包含的代码在默认设置下看起来像这样

<?php
$servername = "localhost";
$username = "root";
$password = "";
$databasename="abc";

// Create connection
$conn = mysqli_connect($servername, $username, $password,$databasename);

// Check connection
if (!$conn) 
    die("Connection failed: " . mysqli_connect_error());

echo "Connected successfully";
?>

我在您的代码中看到的第二件事

$sql =mysqli_query( "DELETE * FROM `products` WHERE `id`=2  LIMIT1 ");

它包含语法错误，应该是

$sql =mysqli_query( $conn,"DELETE FROM `products` WHERE `id`=2  LIMIT 1 ");

【参考方案2】：

Limit 后面有一个空格。

你没有在 mysqli_query() 函数中指定连接。

例如：

<?php
$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno())
  
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  

// Perform queries 
mysqli_query($con,"SELECT * FROM Persons");
mysqli_query($con,"INSERT INTO Persons (FirstName,LastName,Age) 
VALUES ('Glenn','Quagmire',33)");

mysqli_close($con);
?>

你的情况

$sql =mysqli_query( $connectionname,"DELETE * FROM `products` WHERE `id`=2  LIMIT 1 ");

【参考方案3】：

查询错误：$sql =mysqli_query("DELETE * FROM products WHERE id=2 LIMIT1");

将

DELETE * FROM products

替换为

DELETE FROM products

。DELETE 从表中删除行。 类似 mysqli_query 的过程至少需要两个参数

从 mysqli_connect 返回的链接标识符 查询字符串

并且您没有将链接指定为第一个参数，您应该使用返回的链接到 mysqi_query。

$con = mysqli_connect('localhost','root','password','db');
$sql =mysqli_query( $con,"DELETE FROM `products` WHERE `id`=2  LIMIT1 "); 

这个链接帮你链接mysqli_query