在字符串中搜索 char 的所有位置，并将 return 作为逗号分隔的字符串

Question

我有只包含零或一的字符串 (VARCHAR(255))。
我需要搜索所有位置并将它们 return 作为逗号分隔的字符串。我使用 https://dba.stackexchange.com/questions/41961/how-to-find-all-positions-of-a-string-within-another-string

中的解决方案构建了两个查询

到目前为止，这是我的代码：

DECLARE @TERM VARCHAR(5);
SET @TERM = '1';
DECLARE @STRING VARCHAR(255);
SET @STRING = '101011011000000000000000000000000000000000000000';

DECLARE @RESULT VARCHAR(100);
SET @RESULT = '';

SELECT
   @RESULT = @RESULT + CAST(X.pos AS VARCHAR(10)) + ','
FROM
   ( SELECT
      pos = Number - LEN(@TERM)
     FROM
      ( SELECT
         Number
        ,Item = LTRIM(RTRIM(SUBSTRING(@STRING, Number, CHARINDEX(@TERM, @STRING + @TERM, Number) - Number)))
        FROM
         ( SELECT ROW_NUMBER () OVER (ORDER BY [object_id]) FROM sys.all_objects
         ) AS n ( Number )
        WHERE
         Number > 1
         AND Number <= CONVERT(INT, LEN(@STRING))
         AND SUBSTRING(@TERM + @STRING, Number, LEN(@TERM)) = @TERM
      ) AS y
   ) X;

SELECT
   SUBSTRING(@RESULT, 0, LEN(@RESULT));



DECLARE @POS INT;
DECLARE @OLD_POS INT;
DECLARE @POSITIONS VARCHAR(100);
SELECT
   @POSITIONS = '';
SELECT
   @OLD_POS = 0;
SELECT
   @POS = PATINDEX('%1%', @STRING); 
WHILE @POS > 0
   AND @OLD_POS <> @POS
   BEGIN
      SELECT
         @POSITIONS = @POSITIONS + CAST(@POS AS VARCHAR(2)) + ',';
      SELECT
         @OLD_POS = @POS;
      SELECT
         @POS = PATINDEX('%1%', SUBSTRING(@STRING, @POS + 1, LEN(@STRING))) + @POS;
   END;
SELECT
   LEFT(@POSITIONS, LEN(@POSITIONS) - 1);

我想知道这是否可以做到 faster/better？我只搜索单个字符位置，并且我的字符串中只能出现两个字符（0 和 1）。

我已经使用此代码构建了两个函数，运行它们用于 1000 条记录并同时得到相同的结果，所以我无法判断哪个更好。

对于单条记录，第二部分给出 CPU 并且在 Profiler 中读取等于 0，其中第一段代码给我 CPU=16 并且读取=17。

我需要得到如下所示的结果：1,3,5,6,8,9（多次出现时），3 出现一次，NONE 如果没有出现。

Answer 1

一些tallytable和xml的解决方案：

DECLARE @STRING NVARCHAR(100) = '101011011000000000000000000000000000000000000000';

;with cte as(select ROW_NUMBER() OVER(ORDER BY (SELECT NULL)) p 
             from (values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) t1(n) cross join
                  (values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) t2(n) cross join
                  (values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) t3(n))
SELECT STUFF((SELECT  ',' + CAST(p AS VARCHAR(100))
              FROM cte
              WHERE p <= LEN(@STRING) AND SUBSTRING(@STRING, p, 1) = '1'
              FOR XML PATH('')), 1, 1, '')

您只需生成从 1 到 1000 的数字（如果字符串的长度更大，则添加更多连接）并使用 substring 函数过滤所需的值。然后是将行连接到逗号分隔值的标准技巧。

对于旧版本：

;with cte as(SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) p
             FROM sys.all_columns a CROSS JOIN sys.all_columns b)
SELECT STUFF((SELECT  ',' + CAST(p AS VARCHAR(100))
              FROM cte
              WHERE p <= LEN(@STRING) AND SUBSTRING(@STRING, p, 1) = '1'
              FOR XML PATH('')), 1, 1, '')

这里有一篇关于生成范围的好文章http://dwaincsql.com/2014/03/27/tally-tables-in-t-sql/

编辑：

;with cte as(SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) p
               FROM (SELECT 1 AS rn UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 ) t1 CROSS JOIN 
                (SELECT 1 AS rn UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 ) t2 CROSS JOIN 
                (SELECT 1 AS rn UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 ) t3 CROSS JOIN 
                (SELECT 1 AS rn UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 ) t4 CROSS JOIN 
                (SELECT 1 AS rn UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 ) t5 CROSS JOIN 
                (SELECT 1 AS rn UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 UNION ALL SELECT 1 ) t6)

Answer 2

Giorgi 的回答非常聪明，但我更喜欢可读性更强的更老式的方法。我的建议，包括测试用例：

if object_id('UFN_CSVPOSITIONS') is not null
begin
  drop function ufn_csvpositions;
end
go

create function dbo.UFN_CSVPOSITIONS
(
  @string nvarchar(255)
 ,@delimiter nvarchar(1) = ','
)
returns nvarchar(255)
as
begin
  --given a string that contains ones,
  --return a comma-delimited list of the positions of those ones
  --example: '1001' returns '1,4'
  declare @result nvarchar(255) = '';
  declare @i int = 1;
  declare @slen int = len(@string);
  declare @idx int = 0;

  while @i < @slen
  begin
    set @idx = charindex('1',@string,@i);
    if 0 = @idx
    begin
      set @i = @slen;  --no more to be found, break out early
    end
    else
    begin
      set @result = @result + @delimiter + convert(nvarchar(3),@idx);
      set @i = @idx; --jump ahead
    end;
    set @i = @i + 1;
  end  --while

  if (0 < len(@result)) and (',' = substring(@result,1,1))
  begin
    set @result = substring(@result,2,len(@result)-1)
  end

  return @result;
end
go

--test cases
DECLARE @STRING NVARCHAR(255) = '';
set @string = '101011011000000000000000000000000000000000000000';
print dbo.UFN_CSVPOSITIONS(@string,',');
set @string = null;
print dbo.UFN_CSVPOSITIONS(@string,',');
set @string = '';
print dbo.UFN_CSVPOSITIONS(@string,',');
set @string = '1111111111111111111111111111111111111111111111111';
print dbo.UFN_CSVPOSITIONS(@string,',');
set @string = '0000000000000000000000000000000000000000000000000';
print dbo.UFN_CSVPOSITIONS(@string,',');

--lets try a very large # of test cases, see how fast it comes out
--255 "ones" should be the worst case scenario for performance, so lets run through 50k of those.
--on my laptop, here are test case results:
--all 1s   : 13 seconds
--all 0s   :  7 seconds
--all nulls:  1 second
declare @testinput nvarchar(255) = replicate('1',255);
declare @iterations int = 50000;
declare @i int = 0;
while @i < @iterations
begin
  print dbo.ufn_csvpositions(@testinput,',');
  set @i = @i + 1;
end;

--repeat the test using the CTE method.
--the same test cases are as follows on my local:
--all 1s   : 18 seconds
--all 0s   : 15 seconds
--all NULLs: 1  second
set nocount on;
set @i = 0;
set @iterations = 50000;
declare @result nvarchar(255) = '';
set @testinput = replicate('1',255);
while @i < @iterations
begin
  ;with cte as(SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) p
               FROM sys.all_columns a CROSS JOIN sys.all_columns b)
  SELECT @result = STUFF((SELECT  ',' + CAST(p AS VARCHAR(100))
                FROM cte
                WHERE p <= LEN(@testinput) AND SUBSTRING(@testinput, p, 1) = '1'
                FOR XML PATH('')), 1, 1, '')
  print @result;
  set @i = @i + 1;
end;

在字符串中搜索 char 的所有位置，并将 return 作为逗号分隔的字符串

Search all positions of char in string and return as comma separated string

sql

tsql

sql-server

sql-server-2005