Question

我有一张桌子。

l <- list(ab = 1, b = 2, c = 3)
l$a
#[1] 1
l[["a"]]
#NULL
l[["a", exact = FALSE]]
#[1] 1

简单的选择查询无法解决此问题

ID  Contact_Details   Pref
1   1234567890         A 
1   test@test.com      A
2   1234567891         A
2   test1@test.com     B

我想要这样的结果：

select contact_details as mobile,contact_details as email 
from table 
where pref in (A,B);

请注意，我对电子邮件和移动设备的偏好不同，我希望在输出中使用它。

Answer 1

您可以写：

select t1.ID, t1.Contact_Details as Mobile, t1.Pref as PrefMobile,
       t2.Contact_Details as Email, t2.Pref as PrefEmail
from Table t1
inner join Table t2
on t1.ID = t2.ID
where t1.Contact_Details not like '%@%'
and t2.Contact_Details like '%@%'

Answer 2

类似的事情会起作用：

SELECT Id,
       MAX(CASE WHEN Contact_Details NOT LIKE '%@%' THEN Contact_Details END) as Mobile,
       MAX(CASE WHEN Contact_Details NOT LIKE '%@%' THEN Pref END) as MobilePref,
       MAX(CASE WHEN Contact_Details LIKE '%@%' THEN Contact_Details END) as EMAil,
       MAX(CASE WHEN Contact_Details LIKE '%@%' THEN Pref END) as EMailPref
FROM T
GROUP BY ID

Answer 3

    DECLARE @Table TABLE (ID int, [Contact_Details] nvarchar(200), Pref char(1));
    INSERT INTO @Table (ID,[Contact_Details], Pref)
    VALUES 
            (1,'1234567890', 'A') 
           ,(1,'test@test.com', 'A')
           ,(2,'1234567891','A')
           ,(2,'test1@test.com','B');

    WITH Emails
    AS (
    SELECT ID, [Contact_Details], Pref
    FROM 
    @Table
    WHERE CHARINDEX('@',[Contact_Details])>0
    )
    , Mobils
    AS (
    SELECT ID, [Contact_Details], Pref
    FROM 
    @Table
    WHERE CHARINDEX('@',[Contact_Details])<1
    )
    SELECT ISNULL(Mobils.ID,Emails.ID) AS ID, Mobils.[Contact_Details] AS Mobile,Mobils.Pref AS PrefMobile, Emails.[Contact_Details] AS Email,  Emails.Pref AS PrefEmail
    FROM Mobils 
    FULL OUTER JOIN Emails
    ON Mobils.ID=Emails.Id

此查询的结果：

    ID  Mobile  PrefMobile  Email   PrefEmail
    1   1234567890  A   test@test.com   A
    2   1234567891  A   test1@test.com  B

同一行中的数据

3 个答案: