在 PHP 中从 JSON 获取单个项目

Question

我已经搜索了堆栈溢出，但没有找到我想要做什么...所以这里的问题是： 如何仅返回 STEAMID？代码：<?php// define variables and set to empty values$steamID = "";if ($_SERVER["REQUEST_METHOD"] == "POST") {  $steamID = test_input($_POST["steamID"]);}function test_input($data) {  $data = trim($data);  $data = stripslashes($data);  $data = htmlspecialchars($data);  return $data;}$keyApp = '5DE00EE4D07BA728EDAB0B3C2961AA6B';$api = "";$json_data = file_get_contents("http://api.steampowered.com/ISteamUser/GetPlayerSummaries/v0002/?key=$keyApp&steamids=$steamID");$array = json_decode($json_data, true);var_dump($array);?>var_转储：array(1) {  ["response"]=>  array(1) {    ["players"]=>    array(1) {      [0]=>      array(19) {        ["steamid"]=>        string(17) "76561198058543756"        ["communityvisibilitystate"]=>        int(3)        ["profilestate"]=>        int(1)        ["personaname"]=>        string(12) ""Cientistas""        ["commentpermission"]=>        int(1)        ["profileurl"]=>        string(37) "https://steamcommunity.com/id/rhatto/"        ["avatar"]=>        string(116) "https://steamcdn-a.akamaihd.net/steamcommunity/public/images/avatars/c9/c9442f1ea597a7685f8e50f2f7c922321beb4eba.jpg"        ["avatarmedium"]=>        string(123) "https://steamcdn-a.akamaihd.net/steamcommunity/public/images/avatars/c9/c9442f1ea597a7685f8e50f2f7c922321beb4eba_medium.jpg"        ["avatarfull"]=>        string(121) "https://steamcdn-a.akamaihd.net/steamcommunity/public/images/avatars/c9/c9442f1ea597a7685f8e50f2f7c922321beb4eba_full.jpg"        ["avatarhash"]=>        string(40) "c9442f1ea597a7685f8e50f2f7c922321beb4eba"        ["lastlogoff"]=>        int(1593317334)        ["personastate"]=>        int(0)

Answer 1

通过访问唯一的正确路径steamid，显然玩家数组中可能有多个项目。所以你可能需要某种循环。

也许是这样的：

$players = $array['response']['players'];

foreach($players as $player) {

  echo $player['steamid'] . '<br>';

}

或者，如果你的玩家数组中总是有 1 个玩家，你也可以这样做（快速而肮脏）：

echo $array['response']['players'][0]['steamid'];

Answer 2

$steamID = $array['response']['players'][0]['steamid'];