UNNEST/在雪花中展平

Posted 2023-02-23

【中文标题】UNNEST/在雪花中展平

【英文标题】：UNNEST/Flatten in snowflake

【发布时间】：2022-01-21 20:26:50

【问题描述】：

我有一个要转换为雪花的 mysql 查询。

MySQL 查询：

WITH t AS (
    select id,
        date,
        copt,
        split(copt, '|') [ 1 ] as "abc",
        split(copt, '|') [ 2 ] as "def",
        split(copt, '|') [ 3 ] as "xyz",
    from tablename
    where id in (
            123,
            456,
            789,
        )
        and date >= dateadd('day', -6, to_date('2021-12-17'))
        and date <= date '2021-12-17'
        and copt like '%|%|%|%|%|%|%|%|%|%|%'
)
SELECT t.id,
    t.date,
    catId,
    REPLACE(productId, '_', '') as productId,
FROM t
    CROSS JOIN UNNEST(
        split(t."abc", '_'),
        split(t."def", '_'),
        split(t."xyz", '_'),
    ) as x(catId, productId, quantity)
where productid != ''
order by id

我尝试将UNNEST() 替换为FLATTEN()，但没有成功。

谁能帮我把这个查询从 MySQL 转换成 Snowflake？

【参考方案1】：

MySQL 似乎没有 UNNEST 支持，所以我无法阅读相关手册。

您还没有提供任何示例输入和您期望的输出，但假设您 abc、def 和 xzy 是独立大小的数组，并且您想要以下 SQL 应该的所有排列工作：

WITH cte_t(id,date, abc, def, xyz, productid) AS (
    SELECT * FROM VALUES
    (1, '2022-01-23'::date, 'a_b_c_d', 'd_e_f', 'x_y_w', 'not empty'),
    (2, '2022-01-23'::date, 'a_d', 'd_f', 'y_w', 'not empty'),
    (3, '2022-01-23'::date, 'a_c_d', 'f', 'y1_y2_w', 'not empty')
)
SELECT
    t.id,
    t.date,
    a.value AS catid,
    b.value AS productid,
    c.value AS quantity
FROM cte_t AS t
    ,LATERAL SPLIT_TO_TABLE(t.abc, '_') a
    ,LATERAL SPLIT_TO_TABLE(t.def, '_') b 
    ,LATERAL SPLIT_TO_TABLE(t.xyz, '_') c
WHERE productid != ''

给予：

ID	DATE	CATID	PRODUCTID	QUANTITY
1	2022-01-23	a	d	x
1	2022-01-23	b	d	x
1	2022-01-23	c	d	x
1	2022-01-23	d	d	x
1	2022-01-23	a	e	x
1	2022-01-23	b	e	x
1	2022-01-23	c	e	x
1	2022-01-23	d	e	x
1	2022-01-23	a	f	x
1	2022-01-23	b	f	x
1	2022-01-23	c	f	x
1	2022-01-23	d	f	x
1	2022-01-23	a	d	y
1	2022-01-23	b	d	y
1	2022-01-23	c	d	y
1	2022-01-23	d	d	y
1	2022-01-23	a	e	y
1	2022-01-23	b	e	y
1	2022-01-23	c	e	y
1	2022-01-23	d	e	y
1	2022-01-23	a	f	y
1	2022-01-23	b	f	y
1	2022-01-23	c	f	y
1	2022-01-23	d	f	y
1	2022-01-23	a	d	w
1	2022-01-23	b	d	w
1	2022-01-23	c	d	w
1	2022-01-23	d	d	w
1	2022-01-23	a	e	w
1	2022-01-23	b	e	w
1	2022-01-23	c	e	w
1	2022-01-23	d	e	w
1	2022-01-23	a	f	w
1	2022-01-23	b	f	w
1	2022-01-23	c	f	w
1	2022-01-23	d	f	w
2	2022-01-23	a	d	y
2	2022-01-23	d	d	y
2	2022-01-23	a	f	y
2	2022-01-23	d	f	y
2	2022-01-23	a	d	w
2	2022-01-23	d	d	w
2	2022-01-23	a	f	w
2	2022-01-23	d	f	w
3	2022-01-23	a	f	y1
3	2022-01-23	c	f	y1
3	2022-01-23	d	f	y1
3	2022-01-23	a	f	y2
3	2022-01-23	c	f	y2
3	2022-01-23	d	f	y2
3	2022-01-23	a	f	w
3	2022-01-23	c	f	w
3	2022-01-23	d	f	w

好的，这是我所期望的更多数据，但是你去吧。

如果这三个数组大小相同，那么您可以使用一个SPLIT_TO_TALBE 和两个SPLIT_PART：

WITH cte_t(id,date, abc, def, xyz, productid) AS (
    SELECT * FROM VALUES
    (1, '2022-01-23'::date, 'a_b_c', 'd_e_f', 'x_y_w', 'not empty')
)
SELECT
    t.id,
    t.date,
    a.value AS catid,
    split_part(t.def, '_', a.index) AS productid,
    split_part(t.xyz, '_', a.index) AS quantity
FROM cte_t AS t
    ,LATERAL SPLIT_TO_TABLE(t.abc, '_') a
WHERE productid != ''

给予

ID	DATE	CATID	PRODUCTID	QUANTITY
1	2022-01-23	a	d	x
1	2022-01-23	b	e	y
1	2022-01-23	c	f	w