查询我需要这样的

Question

这是我的查询

SELECT CAL.CALENDAR_NAME,CAL.CALENDAR_ID,CALDAY.CALENDARDAY_DAYID 
FROM lms_calendar AS CAL 
    LEFT JOIN LMS_CALENDARDAY AS CALDAY 
    ON CAL.CALENDAR_ID = CALDAY.CALENDARDAY_CALENDARID 

，我得到这样

CALENDAR_NAME          CALENDAR_ID CALENDARDAY_DAYID 
-------------------------------------------------- ----------- ----------------- 
Test            1   1 
Test            1   2 
Test            1   3 
Test            1   4 
Test            1   6 

结果，但我需要这样的

calendar_name calendar_dayid calendar_dayid calendar_dayid calendar_dayid calendar_dayid 
test    1     2    3    4    

Answer 1

下面是一个查询使用PIVOT运营商

SELECT calendar_name, 
    [1] AS calendar_dayid, 
    [2] AS calendar_dayid, 
    [3] AS calendar_dayid, 
    [4] AS calendar_dayid, 
    [5] AS calendar_dayid 
FROM (
    SELECT CAL.CALENDAR_NAME,CAL.CALENDAR_ID,CALDAY.CALENDARDAY_DAYID 
    FROM lms_calendar AS CAL 
     LEFT JOIN LMS_CALENDARDAY AS CALDAY 
     ON CAL.CALENDAR_ID = CALDAY.CALENDARDAY_CALENDARID 
    ) AS src 
PIVOT (
    MAX(calendarday_dayid) 
    FOR calendarday_dayid IN ([1], [2], [3], [4], [5]) 
    ) AS pvt