MySQL的ORDER BY場加場

Question

我有一個表

id type  left right 
1 featured 1  2 
2 default 3  1 
3 default 5  2 
4 default 2  7 
5 featured 3  4 
6 featured 3  2 
7 day  1  3 
8 default 12  42 

我需要輸出5個id其中type != day和sum(left + right)排序並通過功能排序，默認

首先，需要各方面的功能類型ORDERING by sum(left + right)，比type = dafule ordering by sum(left + right) LIMIT 5

我想什麼：

5, 6, 1, 8, 4 

謝謝！

Answer 1

排序由「精選」未來首先是IF（）的順序由...如果類型是「精選」，然後用1作爲排序依據，否則，使用2.既然你只有特色和默認可用（限制「日」條目）。否則，將改爲一個CASE/WHEN結構考慮其他類型的

select 
     yt.id, 
     yt.type, 
     yt.left + yt.right as LeftPlusRight 
    from 
     YourTable yt 
    where 
     yt.type <> 'day' 
    order by 
     if(yt.type = 'featured', 1, 2), 
     LeftPlusRight DESC 
    limit 5 

Answer 2

select id 
from your_table 
where `type` != 'day' 
order by `type`, sum(left + right) 
group by `type`  
limit 5 

Answer 3

SELECT 
    ID 
FROM 
    yourTable 
WHERE 
    type <> 'day' 
ORDER BY (type = 'featured') DESC, (`left` + `right`) DESC 
LIMIT 5 

上述查詢給你正確的結果，我認爲。

Answer 4

與預期結果：

5,6，1,8，4

你真正想要的ID通過 type遞減由 left sum和 right降序排序，然後

，所以以下查詢可滿足您的需求：

SELECT 
    id 
FROM 
    tlr 
WHERE 
    `type`!='day' 
ORDER BY 
    `type` DESC, `left`+`right` DESC 
LIMIT 5; 

它的工作原理是這樣的：

mysql [localhost] {msandbox} (test) > select * from tlr; 
+----+----------+------+-------+ 
| id | type  | left | right | 
+----+----------+------+-------+ 
| 1 | featured | 1 |  2 | 
| 2 | default | 3 |  1 | 
| 3 | default | 5 |  2 | 
| 4 | default | 2 |  7 | 
| 5 | featured | 3 |  4 | 
| 6 | featured | 3 |  2 | 
| 7 | day  | 1 |  3 | 
| 8 | default | 12 | 42 | 
+----+----------+------+-------+ 
8 rows in set (0.00 sec) 

mysql [localhost] {msandbox} (test) > select id from tlr where `type`!='day' order by type desc, `left`+`right` desc limit 5; 
+----+ 
| id | 
+----+ 
| 5 | 
| 6 | 
| 1 | 
| 8 | 
| 4 | 
+----+ 
5 rows in set (0.00 sec)