我有一个字符串11111122111131111111
我想创建一个 1 连续出现的次数数组,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]
,但如何找到它们在连续模式中出现的次数。
以下示例适用于 BigQuery Standard SQL
#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`
有结果
[
{
"line": "11111122111131111111",
"result": [
"6",
"4",
"7"
]
}
]
实际的RDBMS有点不清楚
在这里,我们使用一个临时计数表(任何足够大小的表都可以(。 然后我们应用一个标准的间隙和岛屿。
例
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
返回
Seq Cnt
1 6
2 4
3 7
;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]