PHP json_decode 不起作用

Question

您好，我想获取 Steam 用户的角色名称我将数据存储在 .json 格式的文件中。

{
"response": {
    "players": [
        {
            "steamid": "76561198137714668",
            "communityvisibilitystate": 3,
            "profilestate": 1,
            "personaname": "UareBugged",
            "lastlogoff": 1418911040,
            "commentpermission": 1,
            "profileurl": "http://steamcommunity.com/id/uarenotbest/",
            "avatar": "http://cdn.akamai.steamstatic.com/steamcommunity/public/images/avatars/da/daece8a16d866ef9bd03ddc4aa365c5862af1c21.jpg",
            "avatarmedium": "http://cdn.akamai.steamstatic.com/steamcommunity/public/images/avatars/da/daece8a16d866ef9bd03ddc4aa365c5862af1c21_medium.jpg",
            "avatarfull": "http://cdn.akamai.steamstatic.com/steamcommunity/public/images/avatars/da/daece8a16d866ef9bd03ddc4aa365c5862af1c21_full.jpg",
            "personastate": 1,
            "realname": "Michal Šlesár",
            "primaryclanid": "103582791436765601",
            "timecreated": 1400861961,
            "personastateflags": 0,
            "loccountrycode": "SK",
            "locstatecode": "08"
        }
    ]

}

我想将角色名设为变量，但它什么也没做，变量为空我认为 json_decode 不起作用，但我真的不知道。

    $pname = json_decode(file_get_contents("http://api.steampowered.com/ISteamUser/GetPlayerSummaries/v002/?key=KEYCONSORED&Steamids={$_SESSION['T2SteamID64']}"));
    echo $pname['response']['players']['personaname'];

echo 是空的

Answer 1

玩家是一个数组:

$pname['response']['players'][0]['personaname'];