请帮助我是SQL新手,面对下面的情况。我使用谷歌并尝试找到解决方案,但失败了。
我有一个名为TEMP的临时表,其中一列名为results和rows,具体取决于csv字符串的长度。 SELECT * FROM #TEMP(临时表)时,它返回如下数据:
结果
88.47,1,263759,10.00 | 303.53,2,264051,13.00 | 147.92,3,264052,6.00 | 43.26,4,268394,10.00 | 127.7,5,269229,4.00 |
请使用以下链接直接从数据库查看结果:
http://design.northdurban.com/DatabaseResult.png
我需要一个解决方案,从现有的临时表中读取这些数据,然后将其插入另一个包含行和列的临时表中,如下面的链接所示:
所需的输出显示在下面的链接中
http://design.northdurban.com/capture.png
请帮助,因为我确信此帖可以帮助许多其他用户,因为我还没有找到任何现有解决方案。
答案 0 :(得分:1)
首先使用demiliter |
DECLARE @str VARCHAR(max)='88.47,1,263759,10.00| 303.53,2,264051,13.00| 147.92,3,264052,6.00| 43.26,4,268394,10.00| 127.7,5,269229,4.00|'
SELECT Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)')))
FROM (SELECT Cast ('<M>' + Replace(@str, '|', '</M><M>') + '</M>' AS XML) AS Data) AS A
CROSS APPLY Data.nodes ('/M') AS Split(a)
然后使用parsename
技巧
SELECT Id,c1,c2,c3
FROM (SELECT Id=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 4), ';', '.'),
C1=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 3), ';', '.'),
c2=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 2), ';', '.'),
c3=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 1), ';', '.')
FROM (SELECT Cast ('<M>' + Replace(@str, '|', '</M><M>') + '</M>' AS XML) AS Data) AS A
CROSS APPLY Data.nodes ('/M') AS Split(a)) a
WHERE id IS NOT NULL
更新:要获得更好的效果,请尝试此操作。
SELECT c1,c2,c3,c4
FROM (SELECT C1=Replace(Parsename(Replace(Replace(col, '.', ';'), ',', '.'), 4), ';', '.'),
C2=Replace(Parsename(Replace(Replace(col, '.', ';'), ',', '.'), 3), ';', '.'),
C3=Replace(Parsename(Replace(Replace(col, '.', ';'), ',', '.'), 2), ';', '.'),
C4=Replace(Parsename(Replace(Replace(col, '.', ';'), ',', '.'), 1), ';', '.')
FROM (SELECT Split.a.value('.', 'VARCHAR(100)') col
FROM (SELECT Cast ('<M>' + Replace(@str, '|', '</M><M>') + '</M>' AS XML) AS Data) AS A
CROSS APPLY Data.nodes ('/M') AS Split(a))v) a
WHERE c1 IS NOT NULL;
Update2:要从表中解析多行,请使用此代码。
包含数据的样本表
create table #test(string varchar(8000))
insert into #test values
('88.47,1,263759,10.00| 303.53,2,264051,13.00| 147.92,3,264052,6.00| 43.26,4,268394,10.00| 127.7,5,269229,4.00|'),
('88.47,1,263759,10.00| 303.53,2,264051,13.00| 147.92,3,264052,6.00| 43.26,4,268394,10.00| 127.7,5,269229,4.00|')
查询
SELECT c1,c2,c3,c4
FROM (SELECT C1=Replace(Parsename(Replace(Replace(col, '.', ';'), ',', '.'), 4), ';', '.'),
C2=Replace(Parsename(Replace(Replace(col, '.', ';'), ',', '.'), 3), ';', '.'),
C3=Replace(Parsename(Replace(Replace(col, '.', ';'), ',', '.'), 2), ';', '.'),
C4=Replace(Parsename(Replace(Replace(col, '.', ';'), ',', '.'), 1), ';', '.')
FROM (SELECT Split.a.value('.', 'VARCHAR(100)') col
FROM (SELECT Cast ('<M>' + Replace(string, '|', '</M><M>') + '</M>' AS XML)
AS Data from #test) AS A
CROSS APPLY Data.nodes ('/M') AS Split(a))v) a
WHERE c1 IS NOT NULL;
答案 1 :(得分:0)
仅当您有4列时才会有效。 在这种情况下,您可以执行以下操作
SELECT REPLACE(PARSENAME(REPLACE(REPLACE(ColumnName, '.', '~'), ',', '.'), 4), '~', '.'),
REPLACE(PARSENAME(REPLACE(REPLACE(ColumnName, '.', '~'), ',', '.'), 3), '~', '.'),
REPLACE(PARSENAME(REPLACE(REPLACE(ColumnName, '.', '~'), ',', '.'), 2), '~', '.'),
REPLACE(PARSENAME(REPLACE(REPLACE(ColumnName, '.', '~'), ',', '.'), 1), '~', '.')
From #TEMP
答案 2 :(得分:0)
您可以编写一个表值函数来解析字符串,如下所示:
CREATE FUNCTION dbo.parseData ( @stringToSplit VARCHAR(MAX) )
RETURNS
@return TABLE (ID int, Column1 real, Column2 int, Column3 int, Column4 real)
AS
BEGIN
DECLARE @char char;
DECLARE @len int = LEN(@stringToSplit);
DECLARE @buffer varchar(50) = '';
DECLARE @field int = 1;
DECLARE @Column1 real
DECLARE @Column2 int
DECLARE @Column3 int
DECLARE @Column4 real
DECLARE @row int = 1
DECLARE @i int = 1;
WHILE @i <= @len BEGIN
SELECT @char = SUBSTRING(@stringToSplit, @i, 1)
IF @char = ','
BEGIN
IF @field = 1
SET @Column1 = CONVERT(real, @buffer);
ELSE IF @field = 2
SET @Column2 = CONVERT(int, @buffer);
ELSE IF @field = 3
SET @Column3 = CONVERT(int, @buffer);
SET @buffer = '';
SET @field = @field + 1
END
ELSE IF @char = '|'
BEGIN
SET @Column4 = CONVERT(real, @buffer);
INSERT INTO @return (ID, Column1, Column2, Column3, Column4)
VALUES (@row, @Column1, @Column2, @Column3, @Column4);
SET @buffer = '';
SET @row = @row + 1
SET @field = 1
END
ELSE
BEGIN
SET @buffer = @buffer + @char
END
SET @i = @i + 1;
END
RETURN
END
GO
然后像这样调用该函数:
SELECT Col1 = '88.47,1,263759,10.00| 303.53,2,264051,13.00| 147.92,3,264052,6.00| 43.26,4,268394,10.00| 127.7,5,269229,4.00|'
INTO #Temp1;
INSERT INTO #Temp1
VALUES ('88.47,1,263759,10.00| 303.53,2,264051,13.00| 147.92,3,264052,6.00| 43.26,4,268394,10.00| 127.7,5,269229,4.00|')
SELECT data.*
INTO #Temp2
FROM #Temp1 CROSS APPLY parseData(#Temp1.Col1) as data
SELECT *
FROM #Temp2
DROP TABLE #Temp1
DROP TABLE #Temp2
性能:
所以我针对NoDisplayName描述的技术运行了这种技术的性能测试。超过10,000次迭代我的技术占用了13,826,而NoDisplayName占用了36,176,所以我只占38%的时间NoDisplayName。
为了测试这个,我使用了Azure数据库并运行了以下脚本。
-- First two queries to check the results are the same.
-- Note the Parsename technique returns strings rather than reals which is why
-- the last column has .00 at the end of the numbers in the Parsename tecnique.
DECLARE @str VARCHAR(max)='88.47,1,263759,10.00| 303.53,2,264051,13.00| 147.92,3,264052,6.00| 43.26,4,268394,10.00| 127.7,5,269229,4.01|'
SELECT c1,c2,c3, c4
FROM (SELECT C1=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 4), ';', '.'),
C2=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 3), ';', '.'),
C3=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 2), ';', '.'),
C4=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 1), ';', '.')
FROM (SELECT Cast ('<M>' + Replace(@str, '|', '</M><M>') + '</M>' AS XML) AS Data) AS A
CROSS APPLY Data.nodes ('/M') AS Split(a)) a
WHERE c1 IS NOT NULL;
SELECT *
FROM dbo.parseData(@str)
GO
-- Now lets time the Parsename method over 10,000 itterations
SET NOCOUNT ON;
DECLARE @str VARCHAR(max)='88.47,1,263759,10.00| 303.53,2,264051,13.00| 147.92,3,264052,6.00| 43.26,4,268394,10.00| 127.7,5,269229,4.00|'
DECLARE @i int = 0
declare @table table (c1 decimal, c2 int, c3 int, c4 decimal)
DECLARE @Start datetime = GETDATE();
while @i < 1000
begin
INSERT INTO @table
SELECT c1,c2,c3, c4
FROM (SELECT C1=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 4), ';', '.'),
C2=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 3), ';', '.'),
C3=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 2), ';', '.'),
C4=Replace(Parsename(Replace(Replace(Rtrim(Ltrim(Split.a.value('.', 'VARCHAR(100)'))), '.', ';'), ',', '.'), 1), ';', '.')
FROM (SELECT Cast ('<M>' + Replace(@str, '|', '</M><M>') + '</M>' AS XML) AS Data) AS A
CROSS APPLY Data.nodes ('/M') AS Split(a)) a
WHERE c1 IS NOT NULL;
DELETE FROM @table;
set @i = @i + 1;
end
DECLARE @End datetime = GETDATE()
PRINT CONVERT(nvarchar(50),@Start,126) + ' - ' + convert(nvarchar(50),@End,126) + ' - ' + convert(nvarchar(50), DATEDIFF(ms, @start, @end))
GO
-- Now the my technique over 10,000 itterations
SET NOCOUNT ON;
DECLARE @str VARCHAR(max)='88.47,1,263759,10.00| 303.53,2,264051,13.00| 147.92,3,264052,6.00| 43.26,4,268394,10.00| 127.7,5,269229,4.00|'
DECLARE @i int = 0
declare @table table (c1 decimal, c2 int, c3 int, c4 decimal)
DECLARE @Start datetime = GETDATE();
while @i < 1000
begin
INSERT INTO @table
SELECT *
FROM dbo.parseData(@str)
DELETE FROM @table;
set @i = @i + 1;
end
DECLARE @End datetime = GETDATE()
PRINT CONVERT(nvarchar(50),@Start,126) + ' - ' + convert(nvarchar(50),@End,126) + ' - ' + convert(nvarchar(50), DATEDIFF(ms, @start, @end))
GO