从字符串中提取子字符串

Question

在google-bigquery中，我需要拉出域**和**之间的字符串，如下例所示 该字符串位于＆＃34; Site_Data＆＃34;列下

有人能帮助我吗？ 10倍！

Answer 1

见下面的例子

  

#standardSQL
WITH yourTable AS (
  SELECT '756-1__6565656565656, tagtype**unmapped,domain**www.sport.com,userarriveddirectly**False' AS Site_Data
)
SELECT 
  REGEXP_EXTRACT(Site_Data, r'domain\*\*(.*)\*\*') AS x,
  Site_Data
FROM yourTable

Answer 2

所有字符串都有这种格式吗？假设您始终需要**分隔符后面的第三个字符串，有几个不同的选项。

1）使用SPLIT，例如：

#standardSQL
WITH SampleData AS (
  SELECT '756-1__67648582789116,tagtype**unmapped,domain**www.sport.com,userarriveddirectly**False' AS site_data
)
SELECT SPLIT(site_data, '**')[OFFSET(2)] AS visit_type
FROM SampleData;

2）使用REGEXP_EXTRACT，例如：

#standardSQL
WITH SampleData AS (
  SELECT '756-1__67648582789116,tagtype**unmapped,domain**www.sport.com,userarriveddirectly**False' AS site_data
)
SELECT REGEXP_EXTRACT(site_data, r'[^\*]+\*\*[^\*]+\*\*([^\*]+)') AS visit_type
FROM SampleData;

更进一步，如果您要拆分域名和到达类型，可以再次使用SPLIT：

#standardSQL
WITH SampleData AS (
  SELECT '756-1__67648582789116,tagtype**unmapped,domain**www.sport.com,userarriveddirectly**False' AS site_data
)
SELECT
  SPLIT(visit_type)[OFFSET(0)] AS domain,
  SPLIT(visit_type)[OFFSET(1)] AS arrival_type
FROM (
  SELECT SPLIT(site_data, '**')[OFFSET(2)] AS visit_type
  FROM SampleData
);