取每個條件

Question

只有一排我有一個表

+-------+-------+--------+ 
    id | myid | userid 
+-------+-------+--------+ 
    1 | 123 | 321 
    2 | 321 | 123 
    3 | 123 | 444 
    4 | 444 | 123 
    5 | 123 | 555 
+-------+-------+--------+ 

現在我想取我哪裏有myid='123'

都行，但你看我有無處不在，我希望獲取有限的所有行1 所以最終我得到

ids=[1,3,5]

我使用它來獲取所有行：

SELECT * FROM table 
WHERE (myid='123' AND userid!='123') 
OR (userid='123' AND myid!='123') 
ORDER BY id DESC 

但是這會取所有ID 我可以這樣說嗎？

(
(myid='123' AND userid!='123') 
OR 
(userid='123' AND myid!='123') 
LIMIT 1 
) 

Answer 1

這有什麼錯

SELECT * 
FROM Table 
WHERE (myid = '123' and userid <> '123') OR (myid <> '123' and userid = '123') 
ORDER BY id DESC 
LIMIT 1 

Answer 2

這可能工作：

SELECT 
    tbl1.id as id, 
    t2.my_id as my_id, 
    t2.userid as userid 
FROM 
    (SELECT 
     max(t1.id) as id, 
    FROM 
     table t1 
    WHERE 
     t1.my_id = 123 or 
     t1.userid = 123 
    GROUP BY 
     if(t1.my_id=123, t1.my_id, t1.userid), 
     if(t1.my_id=123, t1.userid, t1.my_id)) tbl1 
LEFT OUTER JOIN 
    table t2 on t2.id = tbl1.id 

這應該讓你ids=[2,4,5]

Answer 3

也許你需要的MySQL group_contact功能

select group_concat(id SEPARATOR ',') from table where myid = '123' 
union 
select group_concat(id SEPARATOR ',') from table where userid = '123'; 

個

這隻回報2行

Answer 4

你可以正常化的身份識別碼和用戶ID列這樣

select *, 
      case 
       when myid < userid then myid 
       else userid 
      end as u1, 
      case 
      when myid > userid then myid 
       else userid 
      end as u2 
from t 

給

+------+------+--------+------+------+ 
| id | myid | userid | u1 | u2 | 
+------+------+--------+------+------+ 
| 1 | 123 | 321 | 123 | 321 | 
| 2 | 321 | 123 | 123 | 321 | 
| 3 | 123 | 444 | 123 | 444 | 
| 4 | 444 | 123 | 123 | 444 | 
| 5 | 123 | 555 | 123 | 555 | 
+------+------+--------+------+------+ 

然後制定出最大的ID這樣

select max(id) from 
(
select *, 
      case 
       when myid < userid then myid 
       else userid 
      end as u1, 
      case 
      when myid > userid then myid 
       else userid 
      end as u2 
from t 
) s 
group by u1,u2 

到給這個

+---------+ 
| max(id) | 
+---------+ 
|  2 | 
|  4 | 
|  5 | 
+---------+ 

，然後加入t等這樣

select t.* 
from  t 
where id in 
(
select max(id) from 
(
select *, 
      case 
       when myid < userid then myid 
       else userid 
      end as u1, 
      case 
      when myid > userid then myid 
       else userid 
      end as u2 
from t 
) s 
group by u1,u2 
) 

給這個

+------+------+--------+ 
| id | myid | userid | 
+------+------+--------+ 
| 2 | 321 | 123 | 
| 4 | 444 | 123 | 
| 5 | 123 | 555 | 
+------+------+--------+ 

無需任何硬編碼！如果你更喜歡第一個會話，則將最大值改爲最小值。