如何使用纯sql执行此查询?

I have a table below

我下面有一张桌子

 first      second 
-------     ---------- 
100        0 
200        0 
0           400

I want to get below result

我想要低于结果

 first      second      result 
-------     ---------- ---------- 
100        0            100 
200        0            300 
0           400         -100

As you can see that result parameter is sum of previous (first-sum) How can i write such a query ?

正如您所看到的，结果参数是先前(第一和)的和，我如何编写这样的查询?

MYSQL solution is very simple, but simple solutions are looking for Microsoft Sql Server.

MYSQL解决方案非常简单，但是简单的解决方案正在寻找Microsoft Sql Server。

set @result =0; 
select first, second, @result := @result + first - second as result 
from tablo;

results

结果

first  second  result   
100    0       100  
200    0       300  
0      400     -100

2 个解决方案

#1

Here's a version with a common table expression. It also suffers from the lack-of-ordering issue, so I used second, first to get the desired results.

这里有一个具有公共表表达式的版本。它还存在订购不足的问题，因此我使用了second, first来获得所需的结果。

WITH cte as
    (
    select [first], [second], [first] - [second] as result,
        ROW_NUMBER() OVER (ORDER BY second, first) AS sequence
    from tableo
    )

SELECT t.[first], t.[second], SUM(t2.result) AS result
from cte t
JOIN cte t2 on t.sequence >= t2.sequence
GROUP BY t.[first], t.[second]

#2

Your first problem is that you're assuming order where there is none. A query without an order by clause has no guaranteed order. Tables without a clustered index don't have a defined order.

你的第一个问题是你在没有秩序的地方假设秩序。没有order by子句的查询没有保证的顺序。没有聚集索引的表没有定义的顺序。

So, if we fix that and put an identity column on the table so that we do have a well defined order, you can use a recursive CTE do it (in mssql 2005 and newer):

因此，如果我们将其固定并在表上放置一个标识列，这样我们就有了一个定义良好的顺序，您可以使用递归CTE(在mssql 2005和更新版本中):

with running_sum as (
  select
    t.id, t.first, t.second, t.first-t.second as result
  from
    table t where t.id = 1
  UNION ALL
  select
    t.id, t.first, t.second, r.result+t.first-t.second
  from
    table t
    join running_sum r on r.id = t.id - 1
)
select
  *
from
  running_sum
order by
  id

#1