从 Snowflake 中的列修剪值

Question

我有一个名为 File 的列，其值为 'Mens_Purchaser_Segment_Report' 和 'Loyalist_Audience_Segment_Report'。我想捕获词段之前的所有内容。

我使用了查询：

select
TRIM(file,regexp_substr(file, '_Segment_Report.*')) as  new_col

Output:
Mens_Purch
Loyalist_Audi

如何捕获 Segment 之前的所有内容？

已尝试以下但结果相同-->

TRIM(file,regexp_substr(file, 'S.*'))

TRIM(file,regexp_substr(file, '_S.*'))

Answer 1

尝试

使用regexp_replace

select regexp_replace(fld1, 'Segment', '') from (
select 'Mens_Purchaser_Segment_Report and Loyalist_Audience_Segment_Report' fld1 from dual );

Answer 2

您没有指定结尾文本是否始终为 _Segment_Report，您要求的是 _Segment 之前的任何文本。取决于可以使用的各种解决方案，请参见下文。

create or replace table foo(s string) as select * from values 
  ('Mens_Purchaser_Segment_Report'),
  ('Loyalist_Audience_Segment_Report');
  
-- If you know the suffix you want to remove is always exactly '_Segment_Report'
select s, replace(s, '_Segment_Report', '') from foo;

-- If you know the suffix you want to remove starts with '_Segment' but can have something after
--   - approach 1, where we replace the _Segment and anything after it with nothing
select s, regexp_replace(s, '_Segment.*', '') from foo;

--   - approach 2, where we extract things before _Segment
--     Note: it will behave differently if there are many instances of '_Segment'
select s, regexp_substr(s, '(.*)_Segment.*', 1, 1, 'e') from foo;

从 Snowflake 中的列修剪值

Trimming value from a column in Snowflake

regex

sql

snowflake-cloud-data-platform