基於SQL Server中逗號分隔計數的分割字符串

Question

需要SQL Server查詢。下面的例子

1）19003 IH-10西，聖安東尼奧，78257，TX，美國

在第一個例子中有串四個逗號。所以輸出應該是

19003 IH-10 West 

2）雪佛龍管線公司，4800 Fournace廣場，貝萊爾，77401-2324，TX，美國

在第二個例子中有串在五個逗號。所以輸出應該是

Chevron Pipe Line Company, 4800 Fournace Place 

如果有四個逗號，那麼輸出應該是主要的字符串的一個子達1逗號，如果有五個逗號，那麼輸出應該是主要的字符串，最多的一個子第二個逗號。

在輸入中，只有4或5個逗號。

我可以通過使用UDF來達到這個要求。但我需要一個直接的查詢。

Answer 1

不知道這將如何執行。我強烈推薦更好的方式來分割地址並將其存儲爲單獨的字段，但有許多地址解析解決方案可用。

如果這不是一個選項，下面的工作可能會發揮作用，但可能需要根據性能改變。

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

這是我會怎麼做（注意我的代碼中的註釋）：

-- 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; 

結果

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 - 添加性能測試

下面是一個測試h arness我放在一起;注意我的意見。

的樣本數據

-- #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 

性能測試

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 

結果（100,000行測試）

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. 

看起來Joe C的解決方案（解決方案＃1）速度更快。