如何在php中定位嵌套的JSON對象，並對其進行回顯？

Question

好的，所以我認為您只要看一下代碼就可以實現我想做的事情。

//Get JSON text file (Steam API)
$json = file_get_contents('http://store.steampowered.com/api/appdetails?appids=57690');

//Decode JSON
$game_json = json_decode($json, true);

//Target game name and echo it
echo $game_json['name'];

JSON本身按此順序排列（非結構化，非常抱歉）：

{"57690":{"success":true,"data":{"type":"game","name":"Tropico 4: Steam Special Edition"

所以我的目標是""name":"Tropico 4: Steam Special Edition"" ，這是我想在頁面上回顯的內容。 我不確定是否有幫助，但是"name":出現一次，我的代碼中是否需要類似[0]代碼來定位第一個？ 是嵌套阻止了我，還是$game_json['name']; 定位方式不正確？

任何提示和/或幫助將不勝感激。 謝謝。

Answer 1

將來，請使用print_r($game_json)檢查array結構。

<?php
$json = file_get_contents('http://store.steampowered.com/api/appdetails?appids=57690');
$game_json = json_decode($json, true);
echo $game_json['57690']['data']['name'];
//Tropico 4: Steam Special Edition
echo $game_json['57690']['data']['required_age'];
//0
//etc...

Answer 2

<?php

//This is your json string
$string = {"57690":{"success":true,"data":{"type":"game","name":"Tropico 4: Steam Special Edition"...

//Turn JSON string into object
$data = json_decode($string);

//Turn your object into an array (easier to work with in this case)
$data = (Array)$data;

//Get name of item with "57690" key
$name = $data["57690"]->data->name;

//Echo the name
echo($name);

//You can also echo out all names of all items like this:
foreach($data as $key => $item)
{
    echo($item->data->name);
}