如何连续排名

Question

我有一张像

这样的表格

   name          age
   -----------------
   jagga         25 
   lipu          25 
   aswinee       24 
   lisu          23 
   ankita        25 
   jhon          23 

我希望像

一样

   name         age     rank
   -------------------------
   ankita       25      1 
   jagga        25      2  
   lipu         25      3 
   aswinee      24      4 
   jhon         23      5  
   lisu         23      6  

name和age必须有order by条款，但rank将从1开始计算。

Answer 1

我们不知道您使用的是哪个dbms，所以
MySQL

SET @id := 0;
SELECT test.name
,      test.age
,      @id := @id + 1 AS rank
FROM   test
ORDER BY age desc, name;

MS SQL Server /的 Oracle / Postgresql ：

SELECT name
,      age
,      ROW_NUMBER() OVER(ORDER BY age DESC, name) AS rank   
FROM   test
ORDER BY age desc, name

Answer 2

仅提及排名函数“row_number”必须具有ORDER BY子句。我同意horsenoname

select name,age,row_number() over (order by age desc,name) as rank from mytable order by age desc,name

Answer 3

请参阅@ a_horse_with_no_name的评论。

如果您的RDBMS是mysql：

select
    t.name,
    t.age,
    @rank := @rank + 1 as rank
from (
    SELECT name, age
    FROM tbl order by name
) as t,
(select @rank := 0) as r

Answer 4

使用分析函数Rank

SELECT NAME, AGE, RANK() OVER (ORDER BY AGE DESC, NAME) Rank FROM STUDENTS;

Answer 5

选择age，row_number()， age结束（按rank desc排序）为table_name  来自age desc，name

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