Geteilte Zeichenfolge basierend auf der Anzahl der durch Komma getrennten Zeichen in SQL Server

Question

Benötigen Sie eine SQL Server-Abfrage. Unten ist das Beispiel

1) 19003 IH-10 West, San Antonio, 78257, TX, USA

Im ersten Beispiel vier Kommas in der Zeichenfolge sind. So soll eine Ausgabe

19003 IH-10 West 

2) Chevron Pipe Line Unternehmen sein, 4800 Fournace Place, Bellaire, 77401-2324, TX, USA

Im zweiten Beispiel gibt es fünf Kommas in der Zeichenkette. So Ausgabe sollte

Chevron Pipe Line Company, 4800 Fournace Place 

werden, wenn es vier Kommas ist, dann sollte Ausgabe ein Teilhaupt String bis zum ersten Komma, und wenn es fünf Kommata ist, dann sollte Ausgabe ein Teil der Hauptsaite bis zu das 2. Komma.

In der Eingabe gibt es nur 4 oder 5 Kommas.

Ich kann diese Anforderung mit einem UDF erreichen. Aber ich brauche eine direkte Abfrage.

Answer 1

Nicht sicher, wie dies funktionieren würde. Ich empfehle eine bessere Möglichkeit, um Adressen zu trennen und als separate Felder zu speichern, gibt es viele Adressenparsing-Lösungen zur Verfügung.

Wenn dies keine Option ist, würde Folgendes funktionieren, aber abhängig von der Leistung möglicherweise geändert werden müssen.

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

Hier ist, wie ich es tun würde (man beachte die Kommentare in meinem Code):

-- 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; 

Ergebnisse

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 - ADDING LEISTUNGSTEST

Unten ist ein Test h Arness ich zusammen; Notiere meine Kommentare.

Beispieldaten

-- #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 

Leistung Test

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 

Ergebnisse (100.000 Reihentest)

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. 

Es sieht so aus, als wäre Joe C's Lösung (Lösung # 1) schneller.