试图在php中获取非对象json的属性

Question

我正在尝试使此代码正常工作，但我不断收到此错误：

试图获取非对象的属性..

我已经验证了我的 json 对象，但仍然没有运气。我搜索了问题，但没有找到任何解决方案。我的代码的第一部分创建JSON 并打印它：

$response['team']['tid'] = $teamData['tid'];
$response['team']['name'] = $teamData['name'];
$response['team']['wins'] = $teamData['wins'];
$response['team']['group'] = $teamData['tgroup'];
$otherTeamNames = $db->query("SELECT name from teams where tgroup= '$teamData[4]'");

while($values = $otherTeamNames->fetch(PDO::FETCH_ASSOC) ){
    if($values['name'] != $teamData['name'])
    $response['team']['otherteam'][] = $values;

}
$response['success'] = 1;
echo json_encode($response);

第二部分将解码 JSON

$json = file_get_contents('http://localhost/pract/getSingleTeamById.php', false, $context);

$teamInformation = json_decode($json);
$success = $teamInformation->{"success"};

$json解码前的输出：

{"team":{"tid":"2","name":"Italy","wins":"4","group":"D","otherteam":[{"name":"Uruguay"},{"name":"England"}]},"success":1}

任何帮助将不胜感激。

Answer 1

你可以这样试试吗？

代替

$teamInformation = json_decode($json);
$success = $teamInformation->{"success"};

使用

$teamInformation = json_decode($json, true);
$success = $teamInformation["success"];