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Javascript 如何使用Ajax发送JSON中的多数据,以及如何在php中接收数据?_Javascript_Php_Jquery_Ajax_Json - Fatal编程技术网

Javascript 如何使用Ajax发送JSON中的多数据,以及如何在php中接收数据?

Javascript 如何使用Ajax发送JSON中的多数据,以及如何在php中接收数据?,javascript,php,jquery,ajax,json,Javascript,Php,Jquery,Ajax,Json,在.js中,我想使用ajax发送json中的多数据。我在php中接收它们时遇到问题 js: 例如,如果taskid为1,如何在debug.txt中编写信息 handler.php: <?php $temp = $_POST; //also i put $_REQUEST not usefull file_put_contents('D:\debug.txt',$temp); //$temp2 = data I don't know how to do it!

在.js中,我想使用ajax发送json中的多数据。我在php中接收它们时遇到问题

js:

例如,如果taskid为1,如何在debug.txt中编写信息

handler.php:

<?php
    $temp = $_POST;  //also i put $_REQUEST not usefull
    file_put_contents('D:\debug.txt',$temp);
    //$temp2 = data   I don't know how to do it!
    if($temp2['taskId']==1){
        file_put_contents('D:\debug.txt',$temp2['infos']);
?>

Json用PHP对您的帖子进行编码:

  $.ajax({
        type:'POST',
        url: 'handler.php',
        data: {taskId:2 , infos:"blahblah"},
        success: function(response){alert "response";},
        error: function(){alert "error";}
    });
以及:


经过大量研究和咨询我的一位朋友后,我得到了答案: 如果我们想在json中使用ajax发送数据,我们的js应该如下所示:

$.ajax({
    type: 'POST',
    url: 'destination.php',
    data: JSON.stringify({
        id: 1,
        name: 'user'
    }),
    headers:{'content-type': 'application/json'},
    success: function(response){
        //assume we have received data from php in json form:
        response = JSON.parse(response);
        dosomething(response);
    },
    error: function(){ alert("Error in Ajaxing"); }
});
现在在php中:

<? php
$json = file_get_contents('php://input');
$ary = json_decode($json);

$result = dothings($ary);
$result = json_encode($result);    
echo(result);
?>

希望这能帮助别人

什么是打印@AbraCadaver:它返回:Array@Barmar请回答这个问题!这没用,它给了我一个空数组:[]@vincent你怎么看这段代码?
$.ajax({
    type: 'POST',
    url: 'destination.php',
    data: JSON.stringify({
        id: 1,
        name: 'user'
    }),
    headers:{'content-type': 'application/json'},
    success: function(response){
        //assume we have received data from php in json form:
        response = JSON.parse(response);
        dosomething(response);
    },
    error: function(){ alert("Error in Ajaxing"); }
});
<? php
$json = file_get_contents('php://input');
$ary = json_decode($json);

$result = dothings($ary);
$result = json_encode($result);    
echo(result);
?>