Question

我正在尝试ajax方法，以使用php将数据插入数据库。但是只有警报消息有效，并且数据没有插入数据库

我的表单：

function register() {
var team = document.getElementById("team").value;
var m1 = document.getElementById("m1").value;
var m2 = document.getElementById("m2").value;   
var m3 = document.getElementById("m3").value;
var m4 = document.getElementById("m4").value;
var email = document.getElementById("email").value;
var number = document.getElementById("number").value;
// Returns successful data submission message when the entered information is stored in database.
var dataString = 'team=' + team + '&m1=' + m1 + '&m2=' + m2 + '&m3' + m3 + '&m4' + m4 + '&email' + email + '&number' + number;
if (team == '' || m1 == '' || m2 == '' || m3 == '' || m4 == '' || email == '' || number == '') {
    alert("Please Fill All Fields");
} else {
    // AJAX code to submit form.
    $.ajax({
        type: "POST",
        url: "workreg.php",
        data: dataString,
        cache: false,
        success: function(html) {
            alert(html);
        }
    });
}
return false;

调用功能：

<?php
// Fetching Values From URL
$team=$_POST['team'];
$m1=$_POST['m1'];
$m2=$_POST['m2'];
$m3=$_POST['m3'];
$m4=$_POST['m4'];
$email=$_POST['m4'];
$number=$_POST['m4'];
$connection = mysql_connect("localhost", "root", ""); // Establishing Connection with Server..
$db = mysql_select_db("event", $connection); // Selecting Database
if (isset($_POST['team'])) {
    $query = mysql_query("insert into workshop values ('$team', '$m1', '$m2','$m3','$m4','$email','$number')"); //Insert Query
    echo "Form Submitted succesfully";
}
mysql_close($connection); // Connection Closed

}

这是php：

{{1}}

？>

每当我单击注册时，它只是显示警报消息，但我检查了数据库，未插入任何值

Answer 1

您的帖子请求中缺少Content-Type。我只使用PostgreSQL，但是Content-Type是使发帖请求正常工作所必需的，也许您应该检查一下。还要检查是否设置为在连接时自动提交到数据库，否则数据不会添加到数据库。

Answer 2

    $('#register').submit(event => {
        event.preventDefault();

        let team = $('#team').val();
        let m1 = $('#m1').val();
        let m2 = $('#m2').val();
        let m3 = $('#m3').val();
        let m4 = $('#m4').val();
        let email = $('#email').val();
        let number = $('#number').val();

        $.ajax({
            url: "workreg.php",
            method: "POST",
            data: { team, m1, m2, m3, m4, email, number }
                .done(data => {
                    console.log(data)
                })
                .fail(err => console.error(err))
        })
    })

Answer 3

实际上，您只是在检查发布变量test是否存在。您不检查数据库查询是否成功执行。在$query

之后检查当前代码

if(!$query){ 
    echo "Form Submitted succesfully"
} else {
     die('Invalid query: ' . mysql_error()); // show  the error 
}

mysql是不推荐使用的功能，所以我正在使用mysqli，这对您来说也更好。永远不要相信用户的输入，所以我也在使用prepare语句。 您应该始终在寻找更新的视频和文章。

$connection = mysqli_connect("localhost", "root", "","event"); // Establishing Connection with Server..

if (isset($_POST['team'])) {
    $query = "insert into workshop values (?, ?, ?,?,?,?,?)"; //Never trust user's input so use prepare 
    $stmt = mysqli_prepare($connection ,$query) ;
    mysqli_stmt_bind_param($stmt,'ssssssi',$team,$m1,$m2,$m3,$m4,$email,$number) ;
    mysqli_stmt_execute($stmt);
    if( mysqli_stmt_affected_rows($stmt) === 1 ) { //it will be int 1 if one row is inserted
        echo "Form Submitted succesfully" ;
    }else {
        echo mysqli_error($connection)
    }
}

mysqli_close($connection); // Connection Closed

一些未来的来源

How can I prevent SQL injection in PHP?

https://phpdelusions.net/pdo（关于PDO，但您会有所想法的。）

http://php.net/manual/en/mysqli.quickstart.prepared-statements.php

使用onclick使用AJAX-php将数据插入数据库

3 个答案: