我想将 SQL 查询转换为密码。拜托,有什么解决方案可以用密码进行 GROUP BY 吗?
SELECT dt.d_year,
item.i_brand_id brand_id,
item.i_brand brand,
Sum(ss_ext_discount_amt) sum_agg
FROM date_dim dt,
store_sales,
item
WHERE dt.d_date_sk = store_sales.ss_sold_date_sk
AND store_sales.ss_item_sk = item.i_item_sk
AND item.i_manufact_id = 427
AND dt.d_moy = 11
GROUP BY dt.d_year,
item.i_brand,
item.i_brand_id
ORDER BY dt.d_year,
sum_agg DESC,
brand_id;
在 Cypher 中,GROUP BY 由所有聚合函数隐式完成。在 WITH/RETURN 语句中,任何不属于聚合的列都将是 GROUP BY 键。
所以例如在
MATCH (n:Person)
RETURN COUNT(n), n.name, n.age
计数将统计所有具有相同名称和年龄的节点。如果我改为做
MATCH (n:Person)
RETURN COUNT(n), n.name, MIN(n.age), MAX(n.age)
我会统计有多少人有相同的名字,以及这个名字的年龄范围。
只要你有一个有限的预测组(比如
enum
左右)UNWIND RANGE (1, 10) AS i
WITH COLLECT({ Val: i % 3 }) AS Item
RETURN REDUCE (
acc = {},
x IN Item | acc {
.*,
V0: CASE x.Val WHEN 0 THEN coalesce(acc.V0 + 1, 1) ELSE acc.V0 END,
V1: CASE x.Val WHEN 1 THEN coalesce(acc.V1 + 1, 1) ELSE acc.V1 END,
V2: CASE x.Val WHEN 2 THEN coalesce(acc.V2 + 1, 1) ELSE acc.V2 END
}) AS grouping
另一种选择是:
UNWIND RANGE(1,15) AS i
WITH i % 4 AS mod, 1 AS unit
RETURN DISTINCT mod, count(unit) as unit