Question

此JavaScript代码使用POST方法将数据从表单发送到PHP，如果数据为true，则PHP检查数据库。但是我不知道如何将PHP的响应发送给JS，提取成功。有人可以解释一下吗？

JS：

$('#submit-btn').on('click', function() {

  var dataString = 'username=' + document.getElementById('username').value + '&password=' + document.getElementById('password').value + '&rememberMe=' + document.getElementById('rememberMe').value;

  $.ajax({
    type: "POST",
    url: "ajaxsubmit.php",
    data: dataString,
    cache: false,
    success: function(){
      //check if what response is   
    } 
  });

ajaxsubmit.php：

<?php
session_start();

//connect

$username =$_POST['username'];
$password =$_POST['password'];  

$sql = "SELECT * FROM user WHERE email ='$username' OR username ='$username' AND password = '$password'";
$result = mysqli_query($conn, $sql);

if(!$row = mysqli_fetch_assoc($result)){     
    //response error
} else{
     //response success
}

?>

Answer 1

你必须在PHP中回应一些东西才能得到回报：

if(!$row = mysqli_fetch_assoc($result)){
    //response error
    echo 'there is a problem';
} else {    
     //response success
     echo 'yippee!';
}

然后您可以按如下方式记录退货：

$('#submit-btn').on('click', function() {

    var dataString = 'username=' + document.getElementById('username').value + '&password=' + document.getElementById('password').value + '&rememberMe=' + document.getElementById('rememberMe').value;
    $.ajax({
        type: "POST",
        url: "ajaxsubmit.php",
        data: dataString,
        cache: false,
        success: function(data){ // 'data' is the variable holding the return from PHP's echo
        //check if what response is   
           console.log(data);
     } 
});

Make sure to watch the AJAX request / response in the browser's developer tools. Make sure you included the jQuery library in the project. The console will reveal errors. AJAX requires a web server.

警告 Little Bobby说 your script is at risk for SQL Injection Attacks. 了解prepared MySQLi语句。即使escaping the string也不安全！

危险 从不存储纯文本密码！请使用 PHP的built-in functions 来处理密码安全问题。如果您使用的PHP版本低于5.5，则可以使用password_hash() compatibility pack。 没有必要escape passwords 或在散列之前对它们使用任何其他清理机制。这样做更改密码并导致不必要的额外编码。

Answer 2

无论你在php中回应什么，都会被发送回ajax。

if(!$row = mysqli_fetch_assoc($result)){
    echo 0;
}
 else{
     echo 1;
}

success: function(response){
    //check if what response is
    console.log( response );
}

Answer 3

您可以退出带有响应参数的方法。喜欢：

ajaxsubmit.php

if(!$row = mysqli_fetch_assoc($result)){     
    exit('error');  //exit with response 'error'
} else{
     exit('success'); //exit with response 'success'
}

JS

$('#submit-btn').on('click', function() {

  var dataString = 'username=' + document.getElementById('username').value + '&password=' + document.getElementById('password').value + '&rememberMe=' + document.getElementById('rememberMe').value;

  $.ajax({
    type: "POST",
    url: "ajaxsubmit.php",
    data: dataString,
    cache: false,
    success: function(response){
      //check if what response is   
      console.log(response);
    } 
  });

如何从PHP发送响应到Ajax？

3 个答案: