php - 注意:尝试获取非对象错误的属性

Question

我正在尝试从以下位置获取数据:

http://api.convoytrucking.net/api.php?api_key=public&show=player&player_name=Mick_Gibson

但是如果我想用这段代码获取player_name变量:

<?  
$js = file_get_contents('http://api.convoytrucking.net/api.php?api_key=public&show=player&player_name=Mick_Gibson');
  $pjs = json_decode($js);
  var_dump($pjs->{'player_name'});
?>

我收到错误:

Notice: Trying to get property of non-object in **\htdocs\index.php on line 9 + var_dump() returns: NULL

var_dump($pjs) 返回:

array(1) { [0]=> object(stdClass)#52 (15) { ["player_name"]=> string(11) "Mick_Gibson" ["player_id"]=> int(88) ["rank"]=> string(12) "FIRE TURTLEE" ["lastseen"]=> int(1393797692) ["registration_date"]=> string(19) "2012-08-10 17:01:34" ["last_mission_date"]=> string(19) "2014-03-02 21:41:50" ["time_offset"]=> int(1) ["house_id"]=> int(611) ["fines"]=> int(0) ["wanted"]=> int(0) ["police_badge"]=> bool(true) ["vip"]=> bool(false) ["staff"]=> NULL ["stats"]=> object(stdClass)#53 (23) { ["score"]=> int(2941) ["convoy_score"]=> int(818) ["ARTIC"]=> int(515) ["DUMPER"]=> int(565) ["TANKER"]=> int(56) ["CEMENT"]=> int(163) ["TRASH"]=> int(7) ["ARMORED"]=> int(9) ["VAN"]=> int(501) ["TOW"]=> int(502) ["COACH"]=> int(4) ["LIMO"]=> int(97) ["ARRESTS"]=> int(272) ["GTA"]=> int(67) ["BURGLAR"]=> int(122) ["HEIST"]=> int(1) ["PLANE"]=> int(48) ["HELI"]=> int(12) ["FAILED"]=> int(312) ["OVERLOADS"]=> int(160) ["TRUCK_LOADS"]=> int(1275) ["ODOMETER"]=> int(28320798) ["TIME"]=> int(2078450) } ["achievements"]=> array(4) { [0]=> string(20) "Professional Trucker" [1]=> string(13) "Gravel Hauler" [2]=> string(12) "Delivery Boy" [3]=> string(7) "Wrecker" } } }

Answer 1

这是因为 $pjs 是一个单元素对象数组，因此首先应该访问数组元素(它是一个对象)，然后访问其属性。

echo $pjs[0]->player_name;

实际上，您粘贴的转储结果已经很清楚地说明了这一点。