无法从php解析json
我有这样一个JSON:无法从php解析json,php,json,Php,Json,我有这样一个JSON: { "avatar": [ { "Trophies": 1022, "clanLevel": 1, "Attack K Factor": -1921682810, "Attacks Won": 1, "freeGems": -1916837036, "clanBadge": 0, "clanCaslteLevel": 5, "currentHomeId": 1288
{
"avatar": [
{
"Trophies": 1022,
"clanLevel": 1,
"Attack K Factor": -1921682810,
"Attacks Won": 1,
"freeGems": -1916837036,
"clanBadge": 0,
"clanCaslteLevel": 5,
"currentHomeId": 12888426248,
"clanRole": 2,
"exp": 5013,
"homeId": 12888426248,
"Attacks Lost": -1307141699,
"clanId": 326417604098,
"boughtGems": -1517098100,
"userNameChange": false,
"numOfNameChanges": 0,
"level": 111,
"league": 5,
"userName": "King Shiv",
"nameTag": 1440968203000,
"clanName": "lol",
"Defenses Won": 17,
"maxCcTroops": 30,
"gems": -1370568149,
"Defenses Lost": -2055915376,
"townHall": 9,
"inWar": 1,
"Attack Rating": -1000115629
}
]
}
$url = "http://185.112.249.77:9999/Api/Player?player=1;
$url = preg_replace("/ /", "%20", $url);
$jsondata = file_get_contents($url);
$data = json_decode($jsondata, true);
echo "IGN: ".$data['avatar']['userName'];
echo "<br />Town Hall: ".$data['avatar']['townHallLevel'];
echo "<br />Level: ".$data['avatar']['level'];
echo "<br />Trophies: ".$data['avatar']['trophies'];
echo "<br />".$data['avatar']['clanRole'];
我试着这样解析它:
{
"avatar": [
{
"Trophies": 1022,
"clanLevel": 1,
"Attack K Factor": -1921682810,
"Attacks Won": 1,
"freeGems": -1916837036,
"clanBadge": 0,
"clanCaslteLevel": 5,
"currentHomeId": 12888426248,
"clanRole": 2,
"exp": 5013,
"homeId": 12888426248,
"Attacks Lost": -1307141699,
"clanId": 326417604098,
"boughtGems": -1517098100,
"userNameChange": false,
"numOfNameChanges": 0,
"level": 111,
"league": 5,
"userName": "King Shiv",
"nameTag": 1440968203000,
"clanName": "lol",
"Defenses Won": 17,
"maxCcTroops": 30,
"gems": -1370568149,
"Defenses Lost": -2055915376,
"townHall": 9,
"inWar": 1,
"Attack Rating": -1000115629
}
]
}
$url = "http://185.112.249.77:9999/Api/Player?player=1;
$url = preg_replace("/ /", "%20", $url);
$jsondata = file_get_contents($url);
$data = json_decode($jsondata, true);
echo "IGN: ".$data['avatar']['userName'];
echo "<br />Town Hall: ".$data['avatar']['townHallLevel'];
echo "<br />Level: ".$data['avatar']['level'];
echo "<br />Trophies: ".$data['avatar']['trophies'];
echo "<br />".$data['avatar']['clanRole'];
它不返回任何值。为什么会这样
它只是返回:
签名:
大会堂:
级别:
奖杯:使用json\u解码,第二个参数设置为true将返回数组
$data['avatar']有子数组,所以您可以像访问整个子数组一样访问它
$data = json_decode($json,true);
$subarray = $data['avatar'][0];
echo $subarray['userName']; // King Shiv
请参见您的url似乎有问题。使用cURL进行了尝试,但返回的结果是无法连接到主机,主机可能是防火墙,阻止了您的配置。你的json解码很好,解决地址问题,你就会得到内容。这是我正在尝试的页面:@shivanpaw此URL不提供json数据。它返回html内容,那么,你如何解码它们呢?谢谢,我现在得到了:D@sandepsureyou忘了用