将列中的逗号分隔值转换为行

Question

我想将逗号分隔值转换为 Redshift 中的行

例如：

store  |location |products
-----------------------------
1      |New York |fruit, drinks, candy...

期望的输出是：

store  |location | products
------------------------------- 
1      |New York | fruit        
1      |New York | drinks         
1      |New York | candy

是否有任何简单的解决方案可以根据分隔符拆分单词并转换为行？我正在研究这个解决方案，但它还不起作用：https://help.looker.com/hc/en-us/articles/360024266693-Splitting-Strings-into-Rows-in-the-Absence-of-Table-Generating-Functions

如有任何建议，我们将不胜感激。

Answer 1

如果你知道值的最大数量，我想你可以split_part():

select t.store, t.location, split_part(products, ',', n.n) as product
 from t join
      (select 1 as n union all
       select 2 union all
       select 3 union all
       select 4
      ) n
      on split_part(products, ',', n.n) <> '';

您还可以使用：

select t.store, t.location, split_part(products, ',', 1) as product
from t 
union all
select t.store, t.location, split_part(products, ',', 2) as product
from t 
where split_part(products, ',', 2) <> ''
union all
select t.store, t.location, split_part(products, ',', 3) as product
from t 
where split_part(products, ',', 3) <> ''
union all
select t.store, t.location, split_part(products, ',', 4) as product
from t 
where split_part(products, ',', 4) <> ''
union all
. . .

Answer 2

首先，您需要创建一个数字 table，因为加入另一个 table 是 redshift 将一行变成多行的唯一方法（没有展平或取消嵌套功能).

例如，一个 table 有 1024 行，其中的值是 1..1024

然后你可以加入并使用split_part()

SELECT
  yourTable.*,
  numbers.ordinal,
  split_part(your_table.products, ',', numbers.ordinal)  AS product
FROM
  yourTable
INNER JOIN
  numbers
    ON  numbers.ordinal >= 1
    AND numbers.ordinal <= regexp_count(your_table.products, ',') + 1

但是...

Redshift 在预测所需行数方面很糟糕。它将连接整个 1024 行，然后拒绝不匹配的行。

它表现得像狗。

因为设计假设是这样的处理总是在加载到 Redshift 之前完成。

Answer 3

CREATE TABLE temptbl  
(
    store INT,
    location  NVARCHAR(MAX),
    products NVARCHAR(MAX)
)



INSERT temptbl   SELECT 1,  'New York', 'Fruit, drinks, candy'

创建的输出 table 当你被创建时

select * from temptbl


;WITH tmp(store, location, DataItem, products) AS
(
    SELECT
        store,
        location,
        LEFT(products, CHARINDEX(',', products + ',') - 1),
        STUFF(products, 1, CHARINDEX(',', products + ','), '')
    FROM temptbl
    UNION all

    SELECT
        store  ,
        location,
        LEFT(products, CHARINDEX(',', products + ',') - 1),
        STUFF(products, 1, CHARINDEX(',', products + ','), '')
    FROM tmp
    WHERE
        products > ''
)

SELECT
    store,
    location,
    DataItem
FROM tmp

您希望在多行中使用逗号分隔值：运行以上命令后你想要的输出：

希望您找到解决方案:)))

Answer 4

MYSQL is also fine

CREATE TABLE test
SELECT 1 store, 'New York' location, 'fruit,drinks,candy' products;

SELECT store, location, product
FROM test
CROSS JOIN JSON_TABLE(CONCAT('["', REPLACE(products, ',', '","'), '"]'),
                      "$[*]" COLUMNS (product VARCHAR(255) PATH "$")) jsontable

store	location	product
1	New York	fruit
1	New York	drinks
1	New York	candy

db<>fiddle here

Answer 5

在 MySQL 中，这将适用于最多四个逗号分隔值。注意 UNION，而不是 UNION ALL。 Fiddle

SELECT store, location,  
       TRIM(SUBSTRING_INDEX(products, ',', 1)) product
  FROM inventory
 UNION 
SELECT store, location, 
       TRIM(SUBSTRING_INDEX(SUBSTRING_INDEX(products, ',', 2), ',', -1))
  FROM inventory
 UNION 
SELECT store, location, 
       TRIM(SUBSTRING_INDEX(SUBSTRING_INDEX(products, ',', 3), ',', -1))
  FROM inventory
 UNION 
SELECT store, location, 
       TRIM(SUBSTRING_INDEX(SUBSTRING_INDEX(products, ',', 4), ',', -1))
  FROM inventory

我会回应其他人所说的。恕我直言，逗号分隔值是一个糟糕的 table 设计。

丑陋 SQL。能够阅读和推理 SQL 非常重要。清晰总是赢。
而且，AWS 的股东会因此喜欢你，因为你会在 redshift 上花费很多额外的钱。

将列中的逗号分隔值转换为行

Convert comma delimited values in a column into rows

mysql

sql

amazon-redshift

denormalized