我有一张这样的桌子:
type code desc store Sales/Day Stock
-----------------------------------------------
1 AA1 abc 101 3 6
1 AA2 abd 101 4 0
1 AA3 abf 101 4 3
2 BA1 bba 101 5 1
2 BA2 bbc 101 2 1
1 AA1 abc 102 1 4
1 AA2 abd 102 2 0
2 BA1 bba 102 4 2
2 BA2 bbc 102 5 5
etc.
我如何显示这样的结果表:
type code desc Store 101 Store 102
Sales/Day | Stock Sales/Day | Stock
--------------------------------------------------------------
1 AA1 abc 3 6 1 4
1 AA2 abd 4 0 2 0
1 AA3 abf 4 3 0 0
2 BA1 bba 5 1 4 2
2 BA2 bbc 2 1 5 5
etc.
注意: Colspan仅显示。
答案 0 :(得分:0)
第一种方式:FILTER
SELECT
type,
code,
"desc",
COALESCE(SUM(sales_day) FILTER (WHERE store = 101)) as sales_day_101,
COALESCE(SUM(stock) FILTER (WHERE store = 101), 0) as stock_101,
COALESCE(SUM(sales_day) FILTER (WHERE store = 102), 0) as sales_day_102,
COALESCE(SUM(stock) FILTER (WHERE store = 102), 0) as stock_102
FROM mytable
GROUP BY type, code, "desc"
ORDER BY type, code
汇总您的价值观。我选择了SUM
,但在您的情况下,如果行有不同的行,则许多其他聚合函数都可以做到这一点。 FILTER
仅允许您汇总一个商店。
COALESCE
将避免在一个聚合中不存在任何值(例如存储AA3
中的102
)不使用NULL值。
第二种方式,CASE WHEN
SELECT
type,
code,
"desc",
SUM(CASE WHEN store = 101 THEN sales_day ELSE 0 END) as sales_day_101,
SUM(CASE WHEN store = 101 THEN stock ELSE 0 END) as stock_101,
SUM(CASE WHEN store = 102 THEN sales_day ELSE 0 END) as sales_day_102,
SUM(CASE WHEN store = 102 THEN stock ELSE 0 END) as stock_102
FROM mytable
GROUP BY type, code, "desc"
ORDER BY type, code
想法是一样的,但是更新的FILTER
函数被更常见的CASE
子句代替。
注意,“ desc”是Postgres中的保留字。因此,我强烈建议重命名您的列。