Question

我有这张桌子：

code(integer) |number_of_data(integer)| date (Date)

我需要：

按天分组，仅适用于6月的那个月
为每个代码选择当天number_of_data的总和！= 0
为代码选择n_data = 0

对于我提出的前两点：

select sum(number_of_data) nData, TO_CHAR(date, 'DD') dayOfMonth from T1 
where to_char( date, 'mm') = 6 
 and code <> 0
 group by TO_CHAR(date, 'DD') order by TO_CHAR(date, 'DD');

它给了我这个表结果：

nData | dayOfMonth

这很好，无论如何我缺少要求3，其查询将是相同但具有相反的条件（代码= 0）。有没有办法将其添加到上面的查询，以获得此结果：

nData | nDataZero | dayOfMonth

Answer 1

对MS Sql中的语法提出一些看法。这是我用oracle sql方式解决这个问题的方法：）

SELECT sum(nData) nData , sum(nDataZero) nDataZero, T1 
from (
select sum(number_of_data) nData, 0 nDataZero , TO_CHAR(date, 'DD') dayOfMonth 
from T1 
where to_char( date, 'mm') = 6  and code <> 0  
group by TO_CHAR(date, 'DD') 
order by TO_CHAR(date, 'DD') 
UNION 
select 0 nData, sum(number_of_data) nDataZero , TO_CHAR(date, 'DD') dayOfMonth 
from T1 
where to_char( date, 'mm') = 6 and code == 0 
group by TO_CHAR(date, 'DD') 
order by TO_CHAR(date, 'DD')) 
group by T1;

RGDS

Answer 2

您可以使用CASE将数据分成2个插槽，代码= 0，代码＆lt;＆gt; 0：

select sum(number_of_data) nData, TO_CHAR(date, 'DD') dayOfMonth, CASE WHEN code = 0 THEN 0 ELSE 1 AS x  
from T1 
where to_char( date, 'mm') = 6 
group by TO_CHAR(date, 'DD'), x 
order by TO_CHAR(date, 'DD');

Answer 3

假设每天只有一个CODE = 0条目，那么你可以这样做：

SELECT SUM( CASE CODE WHEN 0 THEN NULL ELSE number_of_data END ) AS nData,
       MAX( CASE CODE WHEN 0 THEN number_of_data END ) AS nDataZero,
       EXTRACT( DAY FROM "Date" ) AS dayOfMonth
FROM   T1 
WHERE  EXTRACT( MONTH FROM "Date" ) = 6 
GROUP BY EXTRACT( DAY FROM "Date" )
ORDER BY EXTRACT( DAY FROM "Date" );

如果有多个条目，那么您需要指定如何处理（例如，如果您想要MAX值的总和，请将SUM更改为CODE = 0 ）。

是否可以将多个select语句映射到列中？

3 个答案: