Question

我希望从这个json_encode响应中获得它在php中的使用和状态。我怎么得到它？请求像json_encode（array（'online'=＆gt; $ status））

  {"online":[{"user":"1004","status":"Unmonitored"},
  {"user":"1005","status":"Unmonitored"},
  {"user":"1006","status":"Unmonitored"},
  {"user":"2501","status":"Unmonitored"},
  {"user":"2502","status":"Unmonitored"},
  {"user":"2503","status":"Unmonitored"},
  {"user":"2504","status":"Unmonitored"}]}

Answer 1

您可以使用json_decode（）函数解码JSON，如下所示：

<?php
$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
(json_decode($json);    
?>

另一个例子（读取特定对象）：

<?php
$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}';
$obj = json_decode($json); //read a specific object
print $obj->{'a'}; // 1
?>

更新我看到您将问题更新为“如何使用AJAX获取响应？”

          //start ajax request
            $.ajax({
                url: "data.json",
                success: function(data) {
                    //data downloaded so we call parseJSON function 
                    //and pass downloaded data
                    var json = $.parseJSON(data);
                        //do what you want to do here

                }
            });

Answer 2

PHP 5.4提供了JSON_PRETTY_PRINT选项，可用于json_encode()调用。

http://php.net/manual/en/function.json-encode.php

<?php


$json_string = json_encode($data, JSON_PRETTY_PRINT);

＆GT;

Answer 3

似乎json_decode是您正在寻找的功能。

<?php
$data = '{"online":[
    {"user":"1004","status":"Unmonitored"},
    {"user":"1005","status":"Unmonitored"},
    {"user":"1006","status":"Unmonitored"},
    {"user":"2501","status":"Unmonitored"},
    {"user":"2502","status":"Unmonitored"},
    {"user":"2503","status":"Unmonitored"},
    {"user":"2504","status":"Unmonitored"}
]}';

// json_decode produces stdClass object
$decoded_std = json_decode($data);
var_dump($decoded_std->online[0]->status); // "Unmonitored"

// json_decode produces associative array
$decoded_array = json_decode($data, true); // note the second param
var_dump($decoded_array['online'][0]['status']); // "Unmonitored"

在此处查看托管的工作示例http://ideone.com/xaCx4E。

如何从json_encode获取数据？如何使用AJAX获取响应？

3 个答案: