Sql 如何查找字符在字符串中出现的连续次数
我有一个字符串111111221111131111111 我想创建一个1连续出现次数的数组,即前6个字符是1->两个2->四个1->一个3->七个1 所以我想要的输出是[6,4,7]Sql 如何查找字符在字符串中出现的连续次数,sql,sql-server,count,google-bigquery,bigquery-standard-sql,Sql,Sql Server,Count,Google Bigquery,Bigquery Standard Sql,我有一个字符串111111221111131111111 我想创建一个1连续出现次数的数组,即前6个字符是1->两个2->四个1->一个3->七个1 所以我想要的输出是[6,4,7] 我知道如何查找字符在字符串中出现的次数,但知道如何查找字符在连续模式中出现的次数。以下示例适用于BigQuery标准SQL ;WITH splitString(val) AS ( -- convert the string to xml, seperating the elements by
我知道如何查找字符在字符串中出现的次数,但知道如何查找字符在连续模式中出现的次数。以下示例适用于BigQuery标准SQL
;WITH splitString(val) AS
(
-- convert the string to xml, seperating the elements by spaces
SELECT CAST('<r><i>' + REPLACE(@string,' ','</i><i>') + '</i></r>' AS XML)
)
SELECT [Key],
COUNT(*) [WordCount]
FROM ( -- select all of the values from the xml created in the cte
SELECT p.value('.','varchar(100)') AS [Key]
FROM splitString
CROSS APPLY val.nodes('//i') t (p)) AS t
GROUP BY [Key]
#standardSQL
WITH `project.dataset.table` AS (
SELECT '11111122111131111111' line
)
SELECT line, ARRAY(SELECT LENGTH(e) FROM UNNEST(REGEXP_EXTRACT_ALL(line, r'1+')) e) result
FROM `project.dataset.table`
结果
[
{
"line": "11111122111131111111",
"result": [
"6",
"4",
"7"
]
}
]
对于实际的RDBMS有点不清楚 这里我们使用一个特别的理货表,任何大小合适的表都可以。然后,我们应用标准间隙和孤岛 范例 返回
请标记您正在使用的数据库。我不知道有哪个数据库真正实现了标准SQL。请提及使用RDBMS
Declare @S varchar(500) = '11111122111131111111'
Declare @C varchar(10) = '1'
Select Seq=Row_Number() over (Order by Seq)
,Cnt=count(*)
From (
Select N
,S = substring(@S,N,1)
,Seq = N - Row_Number() over (Order by N)
From ( Select Top (len(@S))
N=Row_Number() Over (Order By (Select NULL))
From master..spt_values n1
) A
Where substring(@S,N,1)=@C
) A
Group By Seq
Order By Seq
Seq Cnt
1 6
2 4
3 7