通过Ajax

Question

如何通过ajax将数据发送到另一个PHP类中的特定方法？在url值中，我指向了类文件，但在哪里可以指定要使用的方法名？

$.ajax({    

        type:'POST',
        url:'ResortController.php',
        data: vidData, 
        cache:false,
        contentType: false,
        processData: false,
        success:function(data){
            console.log("success");
            console.log(vidData);
            //window.location.reload();         
        },
        error: function(data){
            console.log("error");
        }
    });

Answer 1

传递data:vidData中的数据并在调用控制器后指定您的函数名称。

url = BASE_PATH + 'ResortController/FUNCTION_NAME';
vidData = {id: 123, vidName: "testVideo"};

$.ajax({    
        type:'POST',
        url:url,
        data: vidData, 
        cache:false,
        contentType: false,
        processData: false,
        success:function(data){
            console.log("success");
            console.log(data);
            //window.location.reload();         
        },
        error: function(data){
            console.log("error");
        }
    });

在您的函数中使用$_POST，您将获得$_POST['vidData']中的ajax数据。

此外，您需要在ajax data成功时拨打vidData而不是console.log(data)变量。

Answer 2

您需要有一个服务器端机制来处理如何定向您的请求。大概是你发送请求的网址只有类声明......你需要某种调度程序，否则php不知道该怎么做：

<强> jQuery的：

$.ajax({  
        type:'POST',
        url:'/dispatcher.php',
        data: {
            "data":vidData,
            "class":"ResortController",
            "method":"rMethod"
        },
        cache:false,
        success:function(data){
            console.log("success");
            console.log(vidData);
            //window.location.reload();         
        },
        error: function(data){
            console.log("error");
        }
    });

<强> /dispatcher.php

<?php
// This is dangerous if you have no controls in place to allow/restrict
// a program to run a command
// Anyone could send a cURL request and run an automated class/method through
// this mechanism unless you have some way to restrict that
if(!empty($_POST['class']) && !empty($_POST['method'])) {
    // Here you want to have some way to check that a request is valid and not 
    // made from some outside source to run arbitrary commands
    // I will pretend you have an admin identifier function....
    if(is_admin()) {
        call_user_func_array(array($_POST['class'],$_POST['method']),array('data'=>$_POST['data']));
    }
}