如何返回SQL中的第二個最新記錄？

Question

如果：從表

SELECT * 
FROM Table 
WHERE Date=(SELECT MAX(Date) 
      FROM Table 
      ) 

回報最新的記錄，如何得到第二最新記錄？

Answer 1

SELECT * 
FROM Table 
WHERE Date = (SELECT MAX(Date) 
       FROM Table 
       WHERE Date < (SELECT MAX(Date) 
           FROM Table 
          ) 
      ) ; 

或：

SELECT TOP (1) * 
FROM Table 
WHERE Date < (SELECT MAX(Date) 
       FROM Table 
      ) 
ORDER BY Date DESC ; 

或：

SELECT * 
FROM 
    (SELECT t.* 
     , ROW_NUMBER() OVER(ORDER BY Date DESC) AS RowNumber 
    FROM Table t 
) AS tmp 
WHERE RowNumber = 2 ; 

---

如果Date列具有獨特的價值，所有三個查詢可以得到同樣的結果。如果該欄可以有重複的日期，那麼它們可能會給出不同的結果（當第一或第二位有聯繫時）。如果在第二位有聯繫，第一個查詢甚至會給出結果中的多行。

Answer 2

"select TOP (1) * 
from Table 
WHERE Date<(SELECT MAX(Date) FROM Table) 
ORDER BY Date DESC" 

應該這樣做。

Answer 3

請檢查這個代碼。

SELECT * FROM category WHERE Created_Time <(SELECT MAX(Created_Time) FROM category) ORDER BY Created_Time DESC LIMIT 1 

Prasad。在SQL

Answer 4

選擇倒數第二個日期：

SELECT MAX(YourDateColumn) FROM YourTable where ColumnId=2 and YourDateColumn < 
(SELECT MAX(YourDateColumn) FROM YourTable where ColumnId=2) 

希望可以幫助別人。

Answer 5

按日期倒序排列試試這個

SELECT * 
FROM Table 
WHERE Date = (SELECT MAX(Date) FROM Table 
      WHERE Date < (SELECT MAX(Date) FROM Table) 
     ) ;