如何从PHP的API数组中获取单个数据？

Question

我想写一个小的Webapp来显示一些数据。数据来自API。显示了来自API的所有数据，但是如果我只想显示特定的数据，则总是会出错。

我还没有使用API​​的丰富经验，但是我完成的所有教程都通过此echo方法显示了数据。

我当前的代码：

<?php 
$user_id = "3bb8d3bc-ab7a-45fb-8154-ed54897b2c4c";
$api_response = file_get_contents('https://r6tab.com/api/player.php?p_id='.$user_id);
$api_response_decoded = json_decode($api_response, true);

//foreach($api_response_decoded as $val) {

    //echo $val, '<br>';

//}

echo $api_response_decoded->p_name;

?>

api_response输出：

"playerfound":true,"social":{"twitter":"","instagram":"","mixer":"","twitch":"","youtube":"","bio":"","esl":"","discord":"","background":"","aliases":"","embed":""},"seasonal":{"current_NA_mmr":0,"last_NA_mmr":0,"last_NA_mmrchange":0,"current_EU_mmr":3831,"last_EU_mmr":3881,"last_EU_mmrchange":-50,"current_AS_mmr":0,"last_AS_mmr":0,"last_AS_mmrchange":0,"total_casualwins":6,"total_casuallosses":5,"total_casualtotal":11,"total_casualkills":28,"total_casualdeaths":18,"total_rankedwins":38,"total_rankedlosses":24,"total_rankedtotal":62,"total_rankedkills":244,"total_rankeddeaths":201,"total_generalwins":44,"total_generallosses":29,"total_generaltotal":73,"total_generalkills":272,"total_generaldeaths":219,"total_totalbulletshits":3851,"total_totalhs":172,"total_totaltimeplayed":0,"bomb_wins":67,"bomb_losses":41,"bomb_total":108,"secure_wins":0,"secure_losses":0,"secure_total":0,"hostage_wins":1,"hostage_losses":0,"hostage_total":1,"favorite_mode":"bomb"},"matches":[{"casual_wlstatus":"won","casual_winslost":"1 Won","casual_datatime":"10\/02\/19","ranked_wlstatus":"won","ranked_winslost":"6 Won, 2 Lost","ranked_datatime":"10\/02\/19","next":"default","db_p_total_casualwins":1,"db_p_total_casuallosses":0,"db_p_total_casualkills":2,"db_p_total_casualdeaths":2,"db_p_total_rankedwins":6,"db_p_total_rankedlosses":2,"db_p_total_rankedkills":23,"db_p_total_rankeddeaths":23,"db_p_total_totalhs":10,"db_p_NA_currentmmr":0,"db_p_EU_currentmmr":3881,"db_p_AS_currentmmr":0,"NA_mmrchange":0,"EU_mmrchange":189,"AS_mmrchange":0},

我的错误消息：

Trying to get property 'p_name' of non-object in C:\xampp\htdocs\R6S-Stats\stats.php on line 12

Answer 1

通过使用true上的json_decode标志，您正在创建关联的数组而不是对象。删除该标志，您将可以使用对象方法：

$api_response_decoded = json_decode($api_response);
echo $api_response_decoded->p_name; //Scenus

如果要将其保留为关联的数组，则可以像这样访问数据：

echo $api_response_decoded['p_name'];

Answer 2

使用json_decode（'dsfsdf'，true）;

  $p_name = $api_response_decoded['p_name'];

使用json_decode（'dsfsdf）;

$p_name = $api_response_decoded->p_name;