Question

这可能是重复的问题，但需要帮助。我是MySQL的新手。

这是我的问题。

我有一个查询来计算纬度和经度之间的距离。根据距离顺序，我必须返回id。

SELECT dlo.id,
       (3959 * acos(cos(radians(12.9)) * cos(radians(y(gproperty))) * cos(radians(x(gproperty)) - radians(77.5)) +sin(radians(12.9)) * sin(radians(y(gproperty))))) AS distance
FROM db1.gfeature dgf,
     db2.loc dlo,
     db2.cust dcu
WHERE gf.o_type = 6
  AND dcu.id = 240
  AND dgf.o_id = dlo.p_id HAVING distance < 20
ORDER BY distance LIMIT 10;

以

返回

+------+-----------------------+
| id   | distance              |
+------+-----------------------+
|  101 | 0.00025714756425665   |
|  199 | 0.10971525612556807   |
|  722 | 0.22772618588406165   |
+------+-----------------------+

但我只需要显示id列。我昨天asked same-question。但现在我使用三个表来获取数据。加入3个表格时很困惑。

有人可以建议我吗？

我试过这种方式

select id from (
  select 
    dlo.id,
    ( 3959 * acos(   cos( radians(12.9) ) 
                   * cos( radians( y(gproperty) ) )
                   * cos( radians( x(gproperty) ) - radians(77.5) )

                   + sin( radians(12.9) )
                   * sin( radians(y(gproperty) ) )
                 )
    ) AS distance 
  from db1.gfeature dgf 
       join db2.cust dcu, db2.loc dlo 
         on dgf.o_type = 6 and dcu.id = 10 and dgf.o_id = dlo.w_id
) t 
where distance < 10 
order by distance 
limit 10;

但在＆＃34; on ...＆＃34;

附近遇到语法错误

Answer 1

也许是这样的。只需使用子查询即可实现。

SELECT S.ID 
FROM
    (SELECT dlo.id,
           (3959 * acos(cos(radians(12.9)) * cos(radians(y(gproperty))) * cos(radians(x(gproperty)) - radians(77.5)) +sin(radians(12.9)) * sin(radians(y(gproperty))))) AS distance
    FROM db1.gfeature dgf, db2.loc dlo , db2.cust dcu
    WHERE gf.o_type = 6 AND dcu.id = 240 AND dgf.o_id = dlo.p_id 
    HAVING distance < 20) S 
ORDER BY S.distance 
LIMIT 10;

在MySQL中隐藏别名列

1 个答案: