php - 位置查找器的MySQL＃1064错误

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我正在尝试通过以下搜索找到最接近用户的10个条目：

  $data_query = mysqli_query($db, "SELECT *
  111.045 * DEGREES(ACOS(COS(RADIANS(latpoint))
             * COS(RADIANS(latitude))
             * COS(RADIANS(longpoint) - RADIANS(longitude))
             + SIN(RADIANS(latpoint))
             * SIN(RADIANS(latitude)))) AS distance_in_km
 FROM merchants
 JOIN (
 SELECT  33.889676  AS latpoint,  151.193024 AS longpoint
) AS p ON 1=1
ORDER BY distance_in_km
LIMIT 15");

但是我收到以下错误：

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '111.045 * DEGREES(ACOS(COS(RADIANS(latpoint))
             * COS(RADIANS(la' at line 2

我尝试将SELECT *更改为SELECT latitude, longitude（使用``，但是Stack Overflow会弄乱语法），但是并没有做很多事情。

我是PHP和MySQL的新手，所以我对这里可能发生的问题感觉很不深入。

干杯:)

最佳答案

如果还希望选择所有内容，请在第一个*之后添加逗号：SELECT *, 111.045 ...

如果不是，则先删除*