1。无效的JSON

Question

您好我想获得Steam用户的个性我在.json格式的文件中存储了数据。

{
"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"
        }
    ]

}

}

我想让personaname变为变量，但它什么都不做，变量是空的我认为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'];

回声是空的

Answer 1

玩家是阵列：

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

Answer 2

这里有几个错误。

让我逐一解释一下如何找到PHP JSON解码/编码的常见错误。

1。无效的JSON

首先，您的JSON无效，最后错过了结尾}。

更新：就在@tftd评论之后我看到你错误地格式化了你的代码，但无论如何，让我解释一下如何找到问题，因为这不应该像在PHP中那样微不足道。其他错误仍然是有效的。

要检查json_decode无效的原因，请使用json_last_error：它会返回错误编号，这意味着：

0 = JSON_ERROR_NONE = "No error has occurred"
1 = JSON_ERROR_DEPTH = "The maximum stack depth has been exceeded"
2 = JSON_ERROR_STATE_MISMATCH  = "Invalid or malformed JSON"
3 = JSON_ERROR_CTRL_CHAR = "Control character error, possibly incorrectly encoded"
4 = JSON_ERROR_SYNTAX = "Syntax error"
5 = JSON_ERROR_UTF8 = "Malformed UTF-8 characters, possibly incorrectly encode"
6 = JSON_ERROR_RECURSION = "One or more recursive references in the value to be encoded"
7 = JSON_ERROR_INF_OR_NAN = "One or more NAN or INF values in the value to be encoded"
8 = JSON_ERROR_UNSUPPORTED_TYPE = "A value of a type that cannot be encoded was given"

在您的情况下，它正在返回4。所以，我在http://jsonlint.com验证了您的JSON，并在最后找到了遗失的}。

2。 json_decode返回对象，而不是数组

如果您想要访问一个$pname数组，则需要将json_decode行放到：

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

请注意true方法的最后一个参数json_decode。根据{{3}}，当true时，返回的对象将被转换为关联数组。

3。玩家是阵列

修正了您的JSON和json_decode调用，我们可以看到players是一个数组。因此，如果您想要阅读第一个播放器，请使用：

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

固定代码

我没有从网址上阅读，因此我使用了documentation：

<?php

$content = <<<EOD
{
"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"
        }
    ]

 }
}
EOD;

$pname = json_decode($content, true);
echo $pname['response']['players'][0]['personaname'];

这将按预期输出UareBugged。

PHP json_decode不起作用

2 个答案:

1。无效的JSON

2。 json_decode返回对象，而不是数组

3。玩家是阵列

固定代码