查詢找到列

Question

我想找到一個列

Answer 1

的第二，3rd..nth最大值你可以列進行排序爲降序格式，然後只獲得第n行的值的第n個最大值。

編輯::

更新爲每個評論的請求。 警告完全未經測試！

SELECT DOB FROM (SELECT DOB FROM USERS ORDER BY DOB DESC) WHERE ROWID = 6 

上面的東西應該適用於Oracle ...您可能必須先讓語法正確！

Answer 2

沒有指定哪個數據庫，MySQL的，你可以做

SELECT column FROM table ORDER BY column DESC LIMIT 7,10; 

將跳過第7，然後讓你接下來的十年最高。

Answer 3

在SQL Server中，只是做：

select distinct top n+1 column from table order by column desc 

然後扔掉的第一個值，如果你不需要它。

Answer 4

同樣的，你可能需要修復您的數據庫，但如果你想在有潛在價值複製的數據集上2個值，你會想要做一組，以及：

SELECT column 
FROM table 
WHERE column IS NOT NULL 
GROUP BY column 
ORDER BY column DESC 
LIMIT 5 OFFSET 2; 

會跳過前兩個，然後會讓你接下來的五個最高。

Answer 5

純SQL（注意：我建議使用特定於您的DBMS的SQL功能，因爲它可能更有效）。這會讓你獲得第n + 1的最大值（最小，翻轉<）。如果您有重複項，請將其計爲COUNT（DISTINCT VALUE）。

select id from table order by id desc limit 4 ; 
+------+ 
| id | 
+------+ 
| 2211 | 
| 2210 | 
| 2209 | 
| 2208 | 
+------+ 


SELECT yourvalue 
    FROM yourtable t1 
WHERE EXISTS(SELECT COUNT(*) 
       FROM yourtable t2 
       WHERE t1.id  <> t2.id 
        AND t1.yourvalue < t2.yourvalue 
       HAVING COUNT(*) = 3) 


+------+ 
| id | 
+------+ 
| 2208 | 
+------+ 

Answer 6

以下是Oracle的一種方法。這個例子獲得了第9個最高值。只需用包含你正在查找的位置的綁定變量替換9即可。

select created from (
    select created from (
     select created from user_objects 
     order by created desc 
     ) 
     where rownum <= 9 
     order by created asc 
    ) 
    where rownum = 1 

如果您想要第n個唯一值，您可以在最內層的查詢塊上添加DISTINCT。

Answer 7

2005 SQL：

SELECT col1 from 
    (select col1, dense_rank(col1) over (order by col1 desc) ranking 
    from t1) subq where ranking between 2 and @n 

Answer 8

另外一個用於Oracle使用分析功能：

select distinct col1 --distinct is required to remove matching value of column 
from 
(select col1, dense_rank() over (order by col1 desc) rnk 
    from tbl 
) 
where rnk = :b1 

Answer 9

尋找答案時自己，這似乎爲SQL Server的工作只是翻出了這個問題2005（來自Blorgbeard's solution派生）：

SELECT MIN(q.col1) FROM (
    SELECT 
     DISTINCT TOP n col1 
     FROM myTable 
     ORDER BY col1 DESC 
) q; 

有效地，這是一個SELECT MIN(q.someCol) FROM someTable q，通過SELECT DISTINCT...查詢檢索表格的前n個。

Answer 10

考慮下面的員工表和一列工資。

 
+------+ 
| Sal | 
+------+ 
| 3500 | 
| 2500 | 
| 2500 | 
| 5500 | 
| 7500 | 
+------+ 

以下查詢將返回第N個最大元素。

select SAL from EMPLOYEE E1 where 
(N - 1) = (select count(distinct(SAL)) 
      from EMPLOYEE E2 
      where E2.SAL > E1.SAL) 

例如，需要第二最大值時，

select SAL from EMPLOYEE E1 where 
    (2 - 1) = (select count(distinct(SAL)) 
       from EMPLOYEE E2 
       where E2.SAL > E1.SAL) 

 
+------+ 
| Sal | 
+------+ 
| 5500 | 
+------+ 

Answer 11

select sal,ename from emp e where 
2=(select count(distinct sal) from emp where e.sal<=emp.sal) or 
3=(select count(distinct sal) from emp where e.sal<=emp.sal) or 
4=(select count(distinct sal) from emp where e.sal<=emp.sal) order by sal desc; 

Answer 12

Select max(sal) 
from table t1 
where N (select max(sal) 
     from table t2 
     where t2.sal > t1.sal) 

要找到第N個最大SAL。

Answer 13

SELECT * FROM tablename 
WHERE columnname<(select max(columnname) from tablename) 
order by columnname desc limit 1 

Answer 14

的MySQL：

select distinct(salary) from employee order by salary desc limit (n-1), 1; 

Answer 15

表員工

salary 
1256 
1256 
2563 
8546 
5645 

您可以通過此查詢

select salary 
from employee 
where salary=(select max(salary) 
       from employee 
       where salary <(select max(salary) from employee)); 

找到第二個最大值您發現該查詢第三最大值

select salary 
from employee 
where salary=(select max(salary) 
       from employee 
       where salary <(select max(salary) 
           from employee 
           where salary <(select max(salary)from employee))); 

Answer 16

答案： 頂部第二：

select * from (select * from deletetable where rownum <=2 order by rownum desc) where rownum <=1 

Answer 17

（表名稱=學生，列名=標記）

select * from(select row_number() over (order by mark desc) as t,mark from student group by mark) as td where t=4 

Answer 18

（表名=學生，的ColumnName =標記）：

select * 
from student 
where mark=(select mark 
      from(select row_number() over (order by mark desc) as t, 
       mark 
       from student group by mark) as td 
      where t=2) 

Answer 19

您可以使用fo找到列的第n個最大值llowing查詢：

SELECT * FROM TableName a WHERE 
    n = (SELECT count(DISTINCT(b.ColumnName)) 
    FROM TableName b WHERE a.ColumnName <=b.ColumnName); 

Answer 20

我認爲下面的查詢將工作在的Oracle SQL恰到好處......我已經測試了它自己..

信息與此相關的查詢：此查詢使用兩個名爲表employee和department與命名員工列：dept_id（通用於員工和部門）name（員工姓名），salary

和列所屬表：dept_id（普通員工表以及），dept_name

SELECT 
    tab.dept_name,MIN(tab.salary) AS Second_Max_Sal FROM (
    SELECT e.name, e.salary, d.dept_name, dense_rank() over (partition BY d.dept_name   ORDER BY e.salary) AS rank FROM department d JOIN employee e USING (dept_id)) tab 
WHERE 
    rank BETWEEN 1 AND 2 
GROUP BY 
    tab.dept_name 

感謝

Answer 21

Select min(fee) 
from fl_FLFee 
where fee in (Select top 4 Fee from fl_FLFee order by 1 desc) 

漲跌家數四連N.

Answer 22

可以簡化這樣

SELECT MIN(Sal) FROM TableName 
WHERE Sal IN 
(SELECT TOP 4 Sal FROM TableName ORDER BY Sal DESC) 

如果薩爾包含重複的值，則使用此

SELECT MIN(Sal) FROM TableName 
WHERE Sal IN 
(SELECT distinct TOP 4 Sal FROM TableName ORDER BY Sal DESC) 

4將第n個值它可以任何最高值如5或6個等。

Answer 23

這是查詢獲得第n最高從colomn把n = 0第二高，n = 1第三等等......

SELECT * FROM TableName 
WHERE ColomnName<(select max(ColomnName) from TableName)-n order by ColomnName desc limit 1; 

Answer 24

簡單的SQL查詢獲取僱員詳細信息誰在表Employee表中有第N MAX Salary。

sql> select * from Employee order by salary desc LIMIT 1 OFFSET <N - 1>; 

考慮表結構爲：

員工（ ID [INT主鍵的auto_increment]， 名稱[VARCHAR（30）]， 薪水[INT]）;

例子：

如果您需要在上表中，然後第3 MAX工資，查詢將是：

sql> select * from Employee order by salary desc LIMIT 1 OFFSET 2; 

同理：

如果你需要8 MAX薪水在上表中查詢即可：

sql> select * from Employee order by salary desc LIMIT 1 OFFSET 7; 

注： 當你有得到第NMAX價值，你應該給OFFSET爲（N - 1）。

像這樣，你可以在薪水升序的情況下做同樣的操作。

Answer 25

select column_name from table_name 
    ordered by column_name desc limit n-1,1; 

，其中n = 1，2，3，...第n個最大值。

Answer 26

在PostgreSQL中，從Employee表中查找第N個最大工資。

SELECT * FROM Employee WHERE salary in 
(SELECT salary FROM Employee ORDER BY salary DESC LIMIT N) 
ORDER BY salary ASC LIMIT 1;