基于SQL Server中逗号分隔计数的分割字符串

Question

需要SQL Server查询。下面的例子

1）19003 IH-10西，圣安东尼奥，78257，TX，美国

在第一个例子中有串四个逗号。所以输出应该是

19003 IH-10 West 

2）雪佛龙管线公司，4800 Fournace广场，贝莱尔，77401-2324，TX，美国

在第二个例子中有串在五个逗号。所以输出应该是

Chevron Pipe Line Company, 4800 Fournace Place 

如果有四个逗号，那么输出应该是主要的字符串的一个子达1逗号，如果有五个逗号，那么输出应该是主要的字符串，最多的一个子第二个逗号。

在输入中，只有4或5个逗号。

我可以通过使用UDF来达到这个要求。但我需要一个直接的查询。

Answer 1

不知道这将如何执行。我强烈推荐更好的方式来分割地址并将其存储为单独的字段，但有许多地址解析解决方案可用。

如果这不是一个选项，下面的工作可能会发挥作用，但可能需要根据性能改变。

Drop Table #Temp 
Create Table #Temp (Field Varchar(8000)) 
Insert #Temp Values ('19003 IH-10 West, San Antonio, 78257, TX, United States') 
Insert #Temp Values ('Chevron Pipe Line Company, 4800 Fournace Place, Bellaire, 77401-2324, TX, United States') 

;With cteFindCommas As 
(
Select CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field)+1)+1)+1)+1) FifthCommaPos, 
     CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field)+1)+1)+1) FourthCommaPos, 
     CharIndex(',', Field) FirstCommaPos, 
     CharIndex(',', Field, CharIndex(',', Field)+1) SecondCommaPos, 
     * 
    From #Temp 
) 
Select Case When FifthCommaPos > 0 Then Substring(Field, 0, SecondCommaPos) 
      Else Substring(Field, 0, FirstCommaPos) End String,    
     * 
    From cteFindCommas 

Answer 2

这是我会怎么做（注意我的代码中的注释）：

-- sample data 
declare @sometable table (someid int identity, someAddress varchar(200)); 
insert @sometable (someAddress) 
values ('19003 IH-10 West, San Antonio, 78257, TX, United States'), 
     ('Chevron Pipe Line Company, 4800 Fournace Place, Bellaire, 77401-2324, TX, US'); 

-- solution 
select 
    someid, 
    someAddress, 
    newAddress = case commaCount when 4 then C4 when 5 then C5 end -- conditions for 4 or 5 commas only 
from 
(
    select 
    someid, 
    someAddress, 
    commaCount = len(v.a) - len(replace(v.a,',','')), -- calculate the number of commas 
    C4 = substring(v.a, 1, charindex(',', v.a)-1), -- grab everything up to the 1st comma 
    C5 = substring(v.a, 1, charindex(',', v.a, charindex(',', v.a)+1)-1) -- everything up to 2nd comma 
    from @sometable t 
    cross apply (values (t.someAddress)) v(a) -- how I avoid repeated references to t.SomeAddress 
) formatStrings; 

结果

someid someAddress            newAddress 
------- ------------------------------------------------------- ---------------------------------------------- 
1  19003 IH-10 West, San Antonio, 78257, TX, United States 19003 IH-10 West 
2  Chevron Pipe Line Company, 4800 Fournace Place...  Chevron Pipe Line Company, 4800 Fournace Place 

UPDATE - 添加性能测试

下面是一个测试h arness我放在一起;注意我的意见。

的样本数据

-- #1: Create Sample Data 
------------------------------------------------------------------------------------------ 
declare @rows int = 100000; 
if object_id('tempdb..#base') is not null drop table #base; 
if object_id('tempdb..#address') is not null drop table #address; 

-- grabbed 50 random addresses from here: https://www.randomlists.com/random-addresses 
-- manually added the first comma (between street address and city) 
select 
    someId  = identity(int,1,1), 
    someAddress = stuff(addr,patindex('%'+replicate('[0-9]',5)+'%',addr),0,',') 
into #base from (values 
('886 Hartford Ave., Gwynn Oak, MD 21207'), 
('322 Wakehurst St., Deerfield, IL 60015'), 
('62 South Oak Valley St., Lorain, OH 44052'), 
('72 53rd St., New Bern, NC 28560'), 
('569 Swanson Ave., Snellville, GA 30039'), 
('15 Walnut St., New Bern, NC 28560'), 
('94 Kingston St., North Royalton, OH 44133'), 
('77 Rock Creek St., Ocean Springs, MS 39564'), 
('688 S. Bellevue St., Mableton, GA 30126'), 
('61 Queen Rd., Potomac, MD 20854'), 
('72 Jockey Hollow Drive, Elgin, IL 60120'), 
('777 School St., Clarksville, TN 37040'), 
('50 North 1st Street, Mount Prospect, IL 60056'), 
('8004 Valley Drive, Long Beach, NY 11561'), 
('8569 Franklin Court, Lakeland, FL 33801'), 
('837 Buckingham St., Newnan, GA 30263'), 
('46 Birch Hill St., Helena, MT 59601'), 
('617 E. Brookside Drive, Jersey City, NJ 07302'), 
('8133 Valley View St., Clearwater, FL 33756'), 
('42 South Ave., Greensburg, PA 15601'), 
('8782 Oak Meadow St., Helotes, TX 78023'), 
('35 Valley Farms Ave., Racine, WI 53402'), 
('7613 Cobblestone Road, Orlando, FL 32806'), 
('27 Broad Lane, Kaukauna, WI 54130'), 
('9213 Corona Dr., Rockville, MD 20850'), 
('7390 W. Bay Court, Mason, OH 45040'), 
('561 W. St Louis Ave., Silver Spring, MD 20901'), 
('7447 Evergreen Ave., Rocky Mount, NC 27804'), 
('24 NW. Pilgrim Road, Sun Prairie, WI 53590'), 
('846 E. Hall St., Lake Villa, IL 60046'), 
('919 Green Hill Street, New Orleans, LA 70115'), 
('532 Newbridge Lane, Hanover, PA 17331'), 
('3 E. Rose Rd., Waukegan, IL 60085'), 
('15 South Euclid Rd., Springfield Gardens, NY 11413'), 
('453 Mulberry Ave., Parlin, NJ 08859'), 
('8128 New Saddle Court, Fullerton, CA 92831'), 
('9143 Lafayette Ave., Jackson Heights, NY 11372'), 
('481 Edgewater St., Dacula, GA 30019'), 
('8243 Hilltop St., Camp Hill, PA 17011'), 
('70 Lookout St., Marlborough, MA 01752'), 
('9370 South Shirley Drive, King Of Prussia, PA 19406'), 
('8071 Plymouth Road, Huntersville, NC 28078'), 
('593 Charles St., Buckeye, AZ 85326'), 
('9092 Atlantic Ave., Yuma, AZ 85365'), 
('81 Longbranch Road, Ontario, CA 91762'), 
('868 Garfield St., New Lenox, IL 60451'), 
('8333 Kirkland Rd., Plainview, NY 11803'), 
('9714 Prospect Ave., Monroe Township, NJ 08831'), 
('7 N. Atlantic Ave., Reidsville, NC 27320'), 
('9283 Cherry Lane, Waukegan, IL 60085')) a(addr); 

-- index to support large sample data requests: 
create unique clustered index uq_cl_base on #base(someid); 

-- Create randomized addresses: up to 6,250,000 dummy rows (50^4) 
with r(x) as (select top(@rows/50) 1 from #base a, #base b, #base c, #base d), 
base(Field) as 
(
    select Field = 
    max(
     case itemNumber when 1 then substring(item,charindex(' ',item)+1, len(item)+1) end+ 
     case abs(checksum(newid())%10) 
     when 0 then ', Unit '+ cast(abs(checksum(newid())%100)+1 as varchar(3)) 
     when 1 then ', Suite '+ cast(abs(checksum(newid())%100)+1 as varchar(3)) 
     when 2 then ', Penthouse' else '' end)+','+ 
    max(case ItemNumber when 2 then item end)+','+ 
    max(case ItemNumber when 3 then left(item,3)+', ' end)+ 
    max(case ItemNumber when 4 then item end)+', United States' 
    from #base t 
    cross apply dbo.DelimitedSplit8K(t.someAddress,',') 
    group by t.someId 
) 
select addressId = identity(int,1,1), 
     field  = 
     left(cast(abs(checksum(newid())%10000)+1 as varchar(5)), 
      case checksum(newid())%3 when 4 then 4 when 1 then 1 else 3 end)+' '+b.Field 
into #address 
from base b 
cross join r 
order by newid(); 
go 

性能测试

set nocount on; 
print 'solution 1'+char(10)+replicate('-',50); 
go 
declare @st datetime = getdate(), @field varchar(100); 
    ;With cteFindCommas As 
    (
    Select CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field)+1)+1)+1)+1) FifthCommaPos, 
      CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field, CharIndex(',', Field)+1)+1)+1) FourthCommaPos, 
      CharIndex(',', Field) FirstCommaPos, 
      CharIndex(',', Field, CharIndex(',', Field)+1) SecondCommaPos, * 
    From #Address 
) 
    select @field = case when FifthCommaPos > 0 Then Substring(Field, 0, SecondCommaPos) 
        else Substring(Field, 0, FirstCommaPos) end 
    From cteFindCommas; 
print datediff(ms,@st,getdate()); 
go 5 

print 'solution 2'+char(10)+replicate('-',50); 
go 
declare @st datetime = getdate(), @field varchar(100); 
    select @field = case commaCount when 4 then C4 when 5 then C5 end -- conditions for 4 or 5 commas only 
    from 
    (
    select 
     addressId, 
     field, 
     commaCount = len(v.a) - len(replace(v.a,',','')), -- calculate the number of commas 
     C4 = substring(v.a, 1, charindex(',', v.a)-1), -- grab everything up to the 1st comma 
     C5 = substring(v.a, 1, charindex(',', v.a, charindex(',', v.a)+1)-1) -- everything up to 2nd comma 
    from #address t cross apply (values (t.field)) v(a) 
) formatStrings; 
print datediff(ms,@st,getdate()); 
go 5 

结果（100,000行测试）

solution 1 
-------------------------------------------------- 
Beginning execution loop 
133 
133 
130 
126 
126 
Batch execution completed 5 times. 

solution 2 
-------------------------------------------------- 
Beginning execution loop 
156 
160 
156 
157 
153 
Batch execution completed 5 times. 

看起来Joe C的解决方案（解决方案＃1）速度更快。