Question

我有SQL查询：

select title, scale / next_scale, c
from ( select title, scale, scale*D0 AS c, 
              lead(scale) over(partition by title order by scale asc) as next_scale,
              row_number() over(partition by title order by scale asc) as agg_row
       from signatures
     ) agg
where agg_row = 1;

它按预期工作。但是，我真正希望排序“scale”值是几列之间的算术运算，所以我尝试使用AS子句（如上所示）并将查询修改为：

select title, scale / next_scale, c
from ( select title, scale, scale*D0 AS c, 
              lead(scale) over(partition by title order by c asc) as next_scale,
              row_number() over(partition by title order by c asc) as agg_row
       from signatures
     ) agg
where agg_row = 1;

然而，它在ORDER BY c失败了。为什么是这样？我可以替换ORDER BY scale*D0并且它可以正常工作。但是，我最终会想要使用像scale*D0*D1*D2*...*D100这样的术语;而且我不想计算3次不同的时间 - 更不用说查询的物理长度了。我希望有scale*D0*D1*D2*...*D100 AS c然后ORDER BY c。

这可能吗？

我正在使用PostgreSQL。

非常感谢，布雷特

Answer 1

在子查询中计算c：

select title, scale / next_scale, c
from ( select title, scale, c, 
              lead(scale) over(partition by title order by c asc) as next_scale,
              row_number() over(partition by title order by c asc) as agg_row
       from (select title, scale, scale * D0 AS c from signatures) signatures_calc
     ) agg
where agg_row = 1;

Answer 2

您可以按列的序号进行排序：

select title, scale / next_scale, c
from ( select title, scale, scale*D0 AS c, 
          lead(scale) over(partition by title order by scale asc) as next_scale,
          row_number() over(partition by title order by scale asc) as agg_row
   from signatures
 ) agg
where agg_row = 1
order by 3;

Answer 3

当下订单时是子查询，应该填写账单

select title, scale / next_scale, c
from ( select title, scale, scale*D0 AS c, 
          lead(scale) over(partition by title order by scale asc) as next_scale,
          row_number() over(partition by title order by scale asc) as agg_row
       from signatures
       order by 3
 ) agg
where agg_row = 1;

包含AS的SQL SELECT语句

3 个答案: