PHP Mysqli 从数据库中读取对象数组转换为 JSON

Question

$query = "SELECT guid, name, location, description, datetime FROM event";

   $result = doSQLTermError($db_connect, $query);

   while($obj = mysqli_fetch_object($result)){
      $guid = $obj->guid;

      $event_details = new stdClass;

      $event_details->$guid = new stdClass;
          $event_details->$guid->name = $obj->name;
          $event_details->$guid->location = $obj->location;
          $event_details->$guid->description = $obj->description;
          $event_details->$guid->datetime = $obj->datetime;   
   }

   $event = json_encode($event_details);

上面的代码为我提供了以下所需的 JSON 输出：

"data":
{
   "f4ef863ef80b07036de08457ccdf51e3f814":
    {
       "name":"adsg",
       "location":"uhjty;",
       "description":"y;hy",
       "datetime":"2018-01-01 00:00:00"
    }
}

但是它只从数据库中读取最后一个事件记录。如何列出数据库中的所有事件但保持上述 JSON 格式？谢谢。

Answer 1

将 $event_details = new stdClass; 移出 while 循环。
您每次都用新实例覆盖 $event_details 变量，因此您丢失了上一个循环中已经存在的数据。

Answer 2

您必须使用数组从 SQL 查询中获取所有数据：

$query = "SELECT guid, name, location, description, datetime FROM event";

$result = doSQLTermError($db_connect, $query);
$events = [];

while($obj = mysqli_fetch_object($result)){
     $guid = $obj->guid;
     $event_details = new stdClass;
     $event_details->$guid = new stdClass;
     $event_details->$guid->name = $obj->name;
     $event_details->$guid->location = $obj->location;
     $event_details->$guid->description = $obj->description;
     $event_details->$guid->datetime = $obj->datetime;
     $events[] = json_encode($event_details);
}

Answer 3

    $query = "SELECT guid, name, location, description, datetime FROM event";

   $result = doSQLTermError($db_connect, $query);

   $event_details = new stdClass;

   while($obj = mysqli_fetch_object($result)){
      $guid = $obj->guid;

      $event_details->$guid = new stdClass;
          $event_details->$guid->name = $obj->name;
          $event_details->$guid->location = $obj->location;
          $event_details->$guid->description = $obj->description;
          $event_details->$guid->datetime = $obj->datetime;   
   }

   $event = json_encode($event_details);

PHP Mysqli 从数据库中读取对象数组转换为 JSON

PHP Mysqli Reading Array of Objects from Database converting to JSON

php

json

mysqli

object

stdclass