Ajax/JSON请求php处理页面

首先更新您的php文件：1.未付款

//include_once 'db_connect.php';

2. 在这一行之后加上这一行：

   header('Content-Type: application/json');   //this one will tell the server to send a json object back to client (by default it sends text/html)
   

3. 不要使用echo"something"；。相反，在$result变量中获取结果，如下所示：

   $result = array ( 'response' => array('status'=>'error', 'message' => 'password is missing')); //this is just an example for the missing password line 
   

在脚本末尾使用以下内容：

echo json_encode($result);
exit();

第页。S.将php脚本中的response替换为response。

PHP文件应该是这样的：

<?php
include_once 'db_connect.php';
//Start the session if is not started it
//You need it to store here the user's name when he successfully logs in
if (!isset(session_id())) {
    session_start();
}
header("Content-Type: application/json");   //this will tell the browser to send a json object back to client not text/html (as default)
// This function will convert your variable (array) into a JSON object
function result($var){
    echo json_encode($var);
    exit();
}
if(!isset($_GET['submit'])) {
    //check if childsname is empty  
    if(empty($_GET['childsname'])) {
        $response = array('result'=>'fail', 'message' => 'missing childsname');
        result($response);
    }else{
        //Always escape your variables before using them in a database
        //I assumed $mysqli is your database connector
        $childs_name = mysqli_real_escape_string($mysqli, $_GET['childsname']);
    }
    //check if password is empty    
    if(empty($_GET['password'])) {
        $response = array('result'=>'fail', 'message' => 'missing password');
        result($response);
    }else{
        //Always escape your variables before using them in a database
        $password = mysqli_real_escape_string($mysqli, $_GET['password']);
    }

// Query database
$query = "SELECT * FROM app WHERE `child_name` = '$childs_name'";   // You also need to add:    AND `password_column_name` = '$password' LIMIT 1
$results = mysqli_query($mysqli, $query);
// Check if SQL query had erors   
if (!$results){
        $response = array('result'=>'fail', 'message' => 'sql error: ' . mysqli_error($mysqli));
        result($response);
}
// If query was successfull and there are rows do this:
if (mysqli_num_rows($results)>0){ 
    $_SESSION['username'] = $childs_name;
    //You may use this variable to check if the user is logged in like this:
    /*
    if (isset($_SESSION['username'])) {
        // user is logged in
        // do something here
    }
    */
    $response = array('result'=>'success', 'message' => 'user is authenticated'));
    result($response);
} else {
    $response = array('result'=>'fail', 'message' => 'user authentication failed');
    result($response);
}
?>

用这个替换JavaScript代码中的getJSON部分

$.getJSON("includes/login.php",{username:childsname,password:password},function(json)   {
    if(json.result === "success") {
        $("#add_err").html( "Welcome "+childsname+"!");
        //you should redirect to a php page, not a html one AND  on that page you should have a session started ( session_start();   )
        // and there you should check if :
        /*
        if (isset($_SESSION['username'])) {
            // user is logged in
            // do something here
        } else {
            //redirect the user back on this page(login_page)
        }
        */
        //wait two seconds before redirect, otherwise you say 'Welcome kiddo' for nothing because no one will ever see it :)
        setTimeout(function(){
              window.location= "menu.html"; 
        },2000);                          
    }else{
        $("#add_err").html(json.message);
    }
});

稍后更新以解释php:中的$response数组

//$response here is an PHP array with the following structure
$response = array('result'=>'fail', 
                  'message' => 'missing password');

当你呼叫时

结果（$response）；

此数组转换为具有以下结构的JSON对象：

{
     "response":"fail",
     "message": "missing password"
}

因为result（）函数中的最后一条指令是exit（）脚本执行结束，结果在getJSON函数中传递回客户端。

不确定它是否可以作为答案分类，但我想给你一些提示，评论可能太长了。

1） 使用post而不是get。它非常相似，文档中也有很好的例子：https://api.jquery.com/jquery.post/

示例：发布到test.php页面，获取以json格式返回的内容（"John"，"time"=>"2pm"）；？>）。

$.post( "test.php", { func: "getNameAndTime" }, function( data ) {
console.log( data.name ); // John
console.log( data.time ); // 2pm
}, "json");

2） 如何调试？如果您不能直接检查服务器响应，您可以在php端放置一些文件日志来了解那里发生了什么。

3） 在您的php代码中，您会回显"response"而不是"response（响应）"

这可能不能直接解决你的问题，但我希望它能有所帮助。