返回前面带有id的JSON

Question

如何从API中提取每个JSON返回对象前面的id？

电流：

{  
   "1516":{  
      "id":"1516",
      "firstname":"Aluno",
      "lastname":"Teste",
      "email":"teste@gmail.com",
      "dlastaccess":"28-10-2016",
      "coursename":"Curso Demonstra\u00e7\u00e3o"
   }
}

我想如何离开：

[  
   {  
      "id":"1516",
      "firstname":"Aluno",
      "lastname":"Teste",
      "email":"teste@gmail.com",
      "dlastaccess":"28-10-2016",
      "coursename":"Curso Demonstra\u00e7\u00e3o"
   }
]

我正在尝试这样做，因为我的清单模型不能像JSON返回那样工作，所以我无法检查所有复选框。

API：

Answer 1

这就是工作：

$json = '{"1516":{"id":"1516","firstname":"Aluno","lastname":"Teste","email":"teste@gmail.com","dlastaccess":"28-10-2016","coursename":"Curso Demonstra\u00e7\u00e3o"}}';

$values = json_decode($json, true);

$values = array_values($values);

echo json_encode($values);

返回：

[{"id":"1516","firstname":"Aluno","lastname":"Teste","email":"teste@gmail.com","dlastaccess":"28-10-2016","coursename":"Curso Demonstra\u00e7\u00e3o"}]

Answer 2

我猜你的$ result是这种类型的数组（或对象）：

  

$ result [id] = array（检索到sql）;

因为sql会返回结果列表，即使结果是唯一的。想象一下，您的查询返回了两个结果，您如何将它们分开？

尝试：

  

echo json_encode（$ result [$ cursoid]）

或

  

echo json_encode（$ result-＆gt; $ cursoid）

Answer 3

$cursoid = $_GET['idcurso'];

$sql = 'SELECT
user2.id AS ID,
user2.firstname AS Firstname,
user2.lastname AS Lastname,
user2.email AS Email,
IF (user2.lastaccess = 0,"nunca", 
DATE_FORMAT(FROM_UNIXTIME(user2.1astaccess),"%d-96m-W")) AS dLastAccess ,c.fullname AS Coursename
FROM mdl_user_enrolments AS ue
JOIN mdl_enrol AS e ON e.id = ue.enrolid
JOIN mdl_course AS c ON c.id = e.courseid
JOIN mdl_user AS user2 ON user2 .id = ue.userid
WHERE (SELECT timeaccess FROM mdl_user_lastaccess WHERE userid=user2.id AND courseid=c.id) IS NULL and c.id = ?';

$params = array($cursoid);

$result = $DB->get_records_sql($sql, $params);
echo json_encode(array_values($result)); // You can easily access array values without their keys by array_values(array $param) function

Answer 4

使用array_values，默认php函数

$values = $DB->get_records_sql( $sql, $params );
$result = array_values( $values );
return json_encode($result);