从 Bigquery 中的多行字符串中提取多个匹配字符串

Question

我有两个字段， key和一个名为description的多行字符串字段，如下所示：

钥匙	描述
1个	[多行字符串如下]
2个	[多行字符串如下]
3个	[多行字符串如下]

“描述”字段示例：

Some data: 12345
Random string line
System: string that I needA
Some other line here
Some other number of lines here
Some data: qwerty
Random string line
System: string that I needB
Some other line here
Some other number of lines here
Some data: 67890
Random string line
System: string that I needC
Some other line here
Some other number of lines here |

我最终想做的是 output 键的每个“系统”实例：

钥匙	系统
1个	我需要的字符串A
1个	我需要的字符串B
1个	我需要的字符串 C

我曾尝试使用正则表达式，但很难找到下一个换行符，因此尝试从中嵌套，例如

SELECT
*,
SUBSTR(system2, INSTR(system2, 'System: ')+8) AS system2_desc,
FROM (
    SELECT
    *,
    SUBSTR(description, INSTR(description, 'System: ')+8) AS system1_desc,
    FROM (
        SELECT
        CASE WHEN INSTR(description, 'System: ') > 0 THEN 1 ELSE 0 END AS contains_system,
        description,
        FROM my_table
    )
)

然后稍后使用 CHR(10) 查找并删除，但很快这将是不可持续的，并且在不知道系统字符串数量的情况下，我必须考虑比我预期的更多的方式。

是否有 function 可以按照上面的预期 output 提取系统字符串，或者在数组中提取系统字符串，然后我可以对其进行交叉连接？

Answer 1

试试这个：

with mytable as (
select 
    1 as key, 
    """Some data: 12345
Random string line
System: string that I needA
Some other line here
Some other number of lines here
Some data: qwerty
Random string line
System: string that I needB
Some other line here
Some other number of lines here
Some data: 67890
Random string line
System: string that I needC
Some other line here
Some other number of lines here |
""" as description
)
select key, system
from mytable, unnest(REGEXP_EXTRACT_ALL(description, r"System: (.*)\n")) as system 

Answer 2

尝试这个

with mytable as
(
    select """
Some data: 12345
Random string line
System: string that I needA
Some other line here
Some other number of lines here
Some data: qwerty
Random string line
System: string that I needB
Some other line here
Some other number of lines here
Some data: 67890
Random string line
System: string that I needC
Some other line here
Some other number of lines here
""" as descp
),
cte as 
(
    select split(descp,'\n') as ls 
    from mytable
)
select replace(str, "System:", '') from cte,unnest(ls) as str
where starts_with(str,"System:" )