MySQL分组依据，排序/优先级为另一列

我已经研究了这两个问题：

但是，它们都使用聚合函数MAX来获得最高值或填充值，这对我的情况不起作用

为了回答这个问题，我简化了我的情况。以下是我目前的数据：

我想获得每条路线的运营商名称，但与行驶方向有关（即订购或“首选”值）。这是我的伪代码：

if(`direction` = 'west' AND `operatorName` != '') then select `operatorName`
else if(`direction` = 'north' AND `operatorName` != '') then select `operatorName`
else if(`direction` = 'south' AND `operatorName` != '') then select `operatorName`
else if(`direction` = 'east' AND `operatorName` != '') then select `operatorName`

我当前的SQL查询是：

SELECT route, operatorName
FROM test
GROUP BY route

这为我提供了分组，但对于我而言，运算符是错误的：

route | operatorName
--------------------
  95  | James
  96  | Mark
  97  | Justin

我曾尝试应用

ORDER BY

子句，但

GROUP BY

优先。我期望的结果是：

route | operatorName
--------------------
  95  | Richard
  96  | Andrew
  97  | Justin

我不能在这里执行

MAX（）

，因为按字母顺序，“北”在“南”之前。在应用

GROUP BY

子句之前，如何明确说明我的偏好/顺序

还要记住，空字符串不是首选

---

请注意，这是一个简化的示例。实际的查询选择了更多的字段并与其他三个表联接，但查询中没有聚合函数。

您可以使用case构造枚举方向，使它们可以按顺序排序。然后对按路线划分的方向进行排序，然后只选择第一个候选方向

set @c = 1;
set @r = '';

select route
     , direction
     , operatorName
  from (
         select route
              , direction
              , operatorName
              , @c := if (@r = route, @c + 1, 1) as cand
           from (
                  select route
                       , case when direction = 'west'
                              then 1
                              when direction = 'north'
                              then 2
                              when direction = 'south'
                              then 3
                              when direction = 'east'
                              then 4
                              else 5
                         end as enum_direction
                       , direction
                       , operatorName
                )
          order by
                route
              , enum_direction
       )

---

我想出了这个解决方案，然而，它是丑陋的。无论如何，你可以试试：

CREATE TABLE test (
  route INT,
  direction VARCHAR(20),
  operatorName VARCHAR(20)
);

INSERT INTO test VALUES(95, 'east', 'James');
INSERT INTO test VALUES(95, 'west', 'Richard');
INSERT INTO test VALUES(95, 'north', 'Dave');
INSERT INTO test VALUES(95, 'south', 'Devon');
INSERT INTO test VALUES(96, 'east', 'Mark');
INSERT INTO test VALUES(96, 'west', 'Andrew');
INSERT INTO test VALUES(96, 'south', 'Alex');
INSERT INTO test VALUES(96, 'north', 'Ryan');
INSERT INTO test VALUES(97, 'north', 'Justin');
INSERT INTO test VALUES(97, 'south', 'Tyler');

SELECT
    route,
    (SELECT operatorName
       FROM test
     WHERE route = t2.route
       AND direction =
              CASE
                WHEN direction_priority = 1 THEN 'west'
                WHEN direction_priority = 2 THEN 'north'
                WHEN direction_priority = 3 THEN 'south'
                WHEN direction_priority = 4 THEN 'east'
              END) AS operator_name
  FROM (
    SELECT
        route,
        MIN(direction_priority) AS direction_priority
      FROM (
        SELECT
            route,
            operatorName,
            CASE
              WHEN direction = 'west' THEN 1
              WHEN direction = 'north' THEN 2
              WHEN direction = 'south' THEN 3
              WHEN direction = 'east' THEN 4
            END AS direction_priority
          FROM test
      ) t
    GROUP BY route
  ) t2
;

首先，我们选择

方向

更改为数字的所有记录，使其符合要求的顺序。然后，我们对每条路线进行分组，得到最小方向。剩下的内容保留在最外面的查询中-根据找到的最低方向选择运算符名称

输出：

ROUTE OPERATOR_NAME 95 Richard 96 Andrew 97 Justin 路由操作员名称 95理查德 96安德鲁 97贾斯汀 请下次不要将样本数据作为图片，而是作为纯文本或插入内容（最好是）

---

请在

处查看此解决方案。您可以使用MAX示例，您只需“伪造它”。请看这里：

---

没有聚合函数的GROUP BY是任意的。您想做什么（使用GROUP BY）毫无意义@Mihai对于不使用

GROUP BY

来获得我想要的结果的查询，我很乐意接受建议……在我看来，解决方案应该是将表本身和您在4种情况下输入的子查询进行内部联接，但我不确定我能做到。谢谢@Mihai，我已经使用变量而不是窗口函数修改了我的答案。我忘记了它们在mysql中是不可用的。唉，我不能完全肯定这会奏效。请随意编辑。

SELECT *
FROM test
JOIN (SELECT 'west' AS direction, 4 AS weight
      UNION
      SELECT 'north',3
      UNION
      SELECT 'south',2
      UNION
      SELECT 'east',1) AS priority
  ON priority.direction = test.direction
JOIN (
      SELECT route, MAX(weight) AS weight
      FROM test
      JOIN (SELECT 'west' AS direction, 4 AS weight
            UNION
            SELECT 'north',3
            UNION
            SELECT 'south',2
            UNION
            SELECT 'east',1) AS priority
        ON priority.direction = test.direction
      GROUP BY route
) AS t1
  ON t1.route = test.route
    AND t1.weight = priority.weight

select *
from routes r1
where exists (
    select 1
    from routes r2
    where r1.route_id = r2.route_id
    group by r2.route_id
    having min(case r1.direction
                 when 'west' then 1
                 when 'north' then 2
                 when 'south' then 3
                 when 'east' then 4 end) = min(case r2.direction
                                                  when 'west' then 1
                                                  when 'north' then 2
                                                  when 'south' then 3
                                                  when 'east' then 4 end)
)