Question

当我点击.deletePost时，我收到以下错误。似乎$mysqli是undefined，但是我以相同的方式在类似的PHP脚本中使用它并且它没有这个错误，所以我很困惑到底是怎么回事。有人可以解释一下吗？谢谢！

错误：

Notice: Undefined variable: mysqli in C:\wamp\www\NightOwlSoftware\scripts\post_action.php on line 16

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\wamp\www\NightOwlSoftware\scripts\post_action.php on line 16

post_action.php

<?php

include 'db_connect.php';
include 'functions.php';
sec_session_start();
echo "<div>Hello World</div>";
echo "<div>Hello World</div>";
echo "<div>Hello World</div>";
echo "<div>Hello World</div>";
if($_GET['action'] == "deletePost")
        deletePost($_GET['postTitle']);
function deletePost($title){
    $sql = "DELETE FROM blog WHERE Title = '$title'";
    mysqli_query($mysqli, $sql);
}
?>

的functions.php

<?php
function sec_session_start() {
    $session_name = 'sec_session_id'; // Set a custom session name
    $secure = false; // Set to true if using https.
    $httponly = true; // This stops javascript being able to access the session id. 

    ini_set('session.use_only_cookies', 1); // Forces sessions to only use cookies. 
    $cookieParams = session_get_cookie_params(); // Gets current cookies params.
    session_set_cookie_params($cookieParams["lifetime"], $cookieParams["path"], $cookieParams["domain"], $secure, $httponly); 
    session_name($session_name); // Sets the session name to the one set above.
    session_start(); // Start the php session
    session_regenerate_id(); // regenerated the session, delete the old one.  
}
?>

dbconnect.php

<?php
$host="localhost"; // Host name
$username="root"; // username
$password="********"; // password
$dbname="nightowl"; // Database name
$tblname="blog"; // Table name
$mysqli=mysqli_connect($host,$username,$password,$dbname);
mysql_connect("$host", "$username", "$password");
mysql_select_db("$dbname");
?>

的Javascript

$(document).ready(function(){
$('.deletePost').click(function(){
    $.ajax({
        url:"scripts/post_action.php",
        data: {action: "deletePost",  postTitle: $(this).siblings("h3.blog").text()},
        success: function(response){
            $("body").html(response);
            alert('DELETED SUCCESSFULLY');
            }
    });
});
});

Answer 1

那是因为你在一个范围内调用了$ mysqli，它没有定义为$ mysqli不是全局的。您必须将其作为deletePost函数的参数传递。例如：

function deletePost($title, $mysqli){
    $sql = "DELETE FROM blog WHERE Title = '$title'";
    mysqli_query($mysqli, $sql);
}

Answer 2

您不需要mysql_select_db()和mysql_connect()，因为您已经宣布了它。

为什么我的数据库变量在我从Ajax调用的PHP脚本中未定义？

2 个答案: