create table bricks (
brick_id integer,
colour varchar2(10),
shape varchar2(10),
weight integer
);
insert into bricks values ( 1, 'blue', 'cube', 1 );
insert into bricks values ( 2, 'blue', 'pyramid', 2 );
insert into bricks values ( 3, 'red', 'cube', 1 );
insert into bricks values ( 4, 'red', 'cube', 2 );
insert into bricks values ( 5, 'red', 'pyramid', 3 );
insert into bricks values ( 6, 'green', 'pyramid', 1 );
commit;
计算表中的总行数。它返回一行
select count(*) from bricks;
添加 over 子句将其转换为解析。这将保留输入行。
select count(*) over () from bricks;
您查看所有其他列的值，您不能使用 group by
select b.*,
count(*) over () total_count
from bricks b;
分组依据
group by 子句将行拆分为具有相同值的组。例如，以下获取每种颜色的行数和总重量
select colour, count(*), sum ( weight )
from bricks
group by colour;
使用 partition by 子句将输入分割为这样
select b.*,
count(*) over (
partition by colour
) bricks_per_colour,
sum ( weight ) over (
partition by colour
) weight_per_colour
from bricks b;
Order By
按brick_id 对行进行排序。然后显示brick_id小于或等于当前行的行的总行数和重量之和：
select b.*,
count(*) over (
order by brick_id
) running_total,
sum ( weight ) over (
order by brick_id
) running_weight
from bricks b;
分组+ 排序
组合 partition by 和 order by 子句以获得组内的行总数。
按颜色拆分行。然后获取每种颜色的运行行数和行重量
select b.*,
count(*) over (
partition by colour
order by brick_id
) running_total,
sum ( weight ) over (
partition by colour
order by brick_id
) running_weight
from bricks b;
有几行具有相同的重量。因此，当您按此排序时，所有具有相同重量的行都具有相同的运行计数和重量
select b.*,
count(*) over (
order by weight
) running_total,
sum ( weight ) over (
order by weight
) running_weight
from bricks b
order by weight;
select b.*,
count(*) over (
order by weight
rows between unbounded preceding and current row
) running_total,
sum ( weight ) over (
order by weight
rows between unbounded preceding and current row
) running_weight
from bricks b
order by weight;
过滤分析函数
select colour from bricks
group by colour
having count(*) >= 2;
不能在 where 中使用它
select colour from bricks
where count(*) over ( partition by colour ) >= 2;
您必须在子查询中使用分析。然后在外部查询中过滤它
select * from (
select b.*,
count(*) over ( partition by colour ) colour_count
from bricks b
)
where colour_count >= 2;
上一个和下一个值(不显示第一行和不显示最后一行)
select b.*,
lag ( shape ) over ( order by brick_id ) prev_shape,
lead ( shape ) over ( order by brick_id ) next_shape
from bricks b;
使用 first_value 和 last_value 获取有序集合中的第一个或最后一个值
first_value 的结果保持不变。但是对于 last_value 它每行都会改变。这是因为默认窗口子句在当前行停止。
select b.*,
first_value ( weight ) over (
order by brick_id
) first_weight_by_id,
last_value ( weight ) over (
order by brick_id
) last_weight_by_id
from bricks b;
select b.*,
first_value ( weight ) over (
order by brick_id
) first_weight_by_id,
last_value ( weight ) over (
order by brick_id
range between current row and unbounded following
) last_weight_by_id
from bricks b;
