0

我的页面上有一个按钮:

<button class="play"></button>

当我点击它时,它会通过 jQuery 启动视频

$('.play').on('click',function(){
    player.play();
});

但是我还想添加 PHP 代码以在他单击该按钮时将用户 ID 保存在数据库中。

如何混合 PHP 和 jQuery?我不习惯 Ajax,这是解决方案吗?

谢谢你的帮助!

4

2 回答 2

2

添加一个功能:

function add_to_db(user_id){
    $.post('add.php', {userId: user_id}, function(response){
        //do something with the response
        r = JSON.parse(response);
        if (r.status === 'error'){
        //handle the error
        alert(r.message);
        }
    });
});

当你玩时,请执行以下操作

$('.play').on('click',function(){
    player.play();
    user_id = //however you chose to set this in the page
    add_to_db(user_id);
});

你的 add.php:

//perform your db update or insert
//if successful:
$response['status'] = 'success';
//else:
$response['status'] = 'error';
$response['message'] = 'update was not successful';
//look into try/catch blocks for more detailed error messaging.

//then send the response back
echo json_encode($response);
于 2013-09-03T15:49:15.597 回答
0

尝试使用ajax

 $('.play').on('click',function(){
    player.play();
    $.ajax({
      type: "POST",
      url: "some.php",  // your php file name
      data:{id :"id"}, //pass your user id
     success:function(data)
    {
   if(data==true)
    alert("success");
else
  alert("error");
    }
    });
    }

一些.php

<?php
$success=false;  //Declare and initialize a variable 

// do your code to update your database
//and check if the database is updated , if so set $success=true

echo $success;


?>
于 2013-09-03T15:50:39.103 回答