如何选择多列具有相同值的行

Question

I have a sql server table named myTbl like this:

BookID  PageID  textID    Date          Remarks
1       4       9         21-12-2017    1 
2       5       10        15-12-2017    1
3       6       11        13-12-2017    1
4       7       12        11-12-2017    1
2       5       10        22-12-2017    1
4       7       12        18-12-2017    1

I want to group the rows when BookID, PageID and textID have same values and show the result based on the most recent date ascending. For instance 4th & 6th row and 2nd & 5th row.

What I need is:

BookID  PageID  textID    Date           Remarks
1       4       9         21-12-2017     1
2       5       10        15-12-2017     1
2       5       10        22-12-2017     1
3       6       11        13-12-2017     1
4       7       12        11-12-2017     1
4       7       12        18-12-2017     1

How I want to write is:

SELECT *
FROM booksdb.dbo.books 
GROUP BY BookID, PageID, textID Order by Date 

I want all the columns in the table where BookID, PageID, textID are same and arranged by date.

Answer 1

You don't need group by , Because there is not any aggregate function in your query. you can try to use RANK with window function to make row_number then order by the number.

CREATE TABLE T (
   BookID INT,
   PageID  INT,
   textID  INT,
   DATE  DATE,
   Remarks INT
);

INSERT INTO T VALUES (1,4,9 ,'2017-12-21',1);
INSERT INTO T VALUES (2,5,10,'2017-12-15',1);
INSERT INTO T VALUES (2,5,10,'2017-12-22',1);
INSERT INTO T VALUES (3,6,11,'2017-12-13',1);
INSERT INTO T VALUES (4,7,12,'2017-12-11',1);
INSERT INTO T VALUES (4,7,12,'2017-12-18',1);

Query 1 :

;WITH CTE AS (
  SELECT *,RANK() OVER(PARTITION BY BookID, PageID, textID Order by Date) rn
  FROM T
)
SELECT BookID,PageID,textID,[DATE],Remarks
FROM CTE
ORDER BY BookID, PageID, textID,rn

Results :

| BookID | PageID | textID |       DATE | Remarks |
|--------|--------|--------|------------|---------|
|      1 |      4 |      9 | 2017-12-21 |       1 |
|      2 |      5 |     10 | 2017-12-15 |       1 |
|      2 |      5 |     10 | 2017-12-22 |       1 |
|      3 |      6 |     11 | 2017-12-13 |       1 |
|      4 |      7 |     12 | 2017-12-11 |       1 |
|      4 |      7 |     12 | 2017-12-18 |       1 |

Answer 2

All you seem to need is an ORDER BY all the columns you mentioned in the order you mentioned them.

SELECT *
       FROM booksdb.dbo.books 
       ORDER BY bookid,
                pageid,
                textid,
                date;

Edit:

Note: If you want the order of the dates to show the youngest one first (eg 18-12-2017 before 11-12-2017), rather than the other way around, change the ORDER BY clause to:

       ORDER BY bookid,
                pageid,
                textid,
                date DESC;

(Note the DESC after date ).

The desired result may contradict the description you gave and overall they leave room for interpretation regarding this point.