Parse address field to extract City name using t-sql

SQL9 246

Hi All,

I have a table(@Address table) with address field. I would like to extract the city name , state and zip code from it.

I have used STRING_AGG() and STRING_SPLIT() to get the City name, state, zip. In these 3 fields City name is very important to show.

Below is the sample code I am using to get the city, state, & zip but it works fine for city names with space in between them but doesn't work for single word states(top 4 works for states with space but not for states with single word, it pulls other street info).
Also I tried with a table to filter the data based on city names from the table but don't know how to use it correctly with string_split()

Any kind of help is greatly appreciate. I am using 2017 version.

declare @Output as varchar(100) = '100 NORTH MAIN MASON CITY IL 626641104'
SELECT @output = STRING_AGG(cityname, SPACE(1)) FROM (SELECT top 4 ROW_NUMBER() Over (order by (select null)) rn, value FROM STRING_SPLIT(@Output, SPACE(1))
order by rn desc) AS cn (rn,cityname);

SELECT @output = STRING_AGG(cityname, SPACE(1)) FROM (SELECT top 4 ROW_NUMBER() Over (order by (select null)) rn, value FROM STRING_SPLIT(@output, SPACE(1))
order by rn asc) AS Cn (rn,cityname);

SELECT @output = STRING_AGG(cityname, SPACE(1)) FROM (SELECT top 4 ROW_NUMBER() Over (order by (select null)) rn, value FROM STRING_SPLIT(@output, SPACE(1))
/* WHERE value IN (SELECT cityname FROM @City ) */ order by rn desc ) AS Cn (rn,cityname);
select @output

Below is the sample city table data I want to use it to filter the STRING_Split() data

declare @City table (CityName varchar(50))
insert into @City (CityName)
Select 'PLAIN CITY' union all
Select 'MASON CITY' union all
select 'BLOOMINGDALE' union all
select 'CHICAGO'

--select * from @City

declare @Address table (FullAddress varchar(150))
insert into @Address (FullAddress)
Select '5500 North AVENUE, APT. 116 plain city OH 60053' union all
Select '12536 S INDIANA AVE CHICAGO IL 60625' union all
Select 'W MAIN ST PO BOX 111 Addison VA 24293' union all
select '179 ARMY TRAIL RD BLOOMINGDALE IL 60108' union all
select '100 NORTH MAIN MASON CITY IL 626641104'

--select * from @Address

Thanks in advance
RH

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MelissaMa-msft 24,246 Reputation points Moderator

2021-06-01T01:02:41.87+00:00

Hi @SQL9 ,

Could you please validate all the answers and provide any update?

If all are not working or helpful, please provide more sample data and expected output.

Please remember to accept the answers if they helped. Your action would be helpful to other users who encounter the same issue and read this thread.

Thank you for understanding!

Best regards,
Melissa
SQL9 246 Reputation points

2021-06-01T15:22:34.113+00:00

Hi Melissa,
Thanks for taking time and helping on this. The query which you have provided is almost giving same output as mine(I have posted mine). I have used String_Split() and String_Agg() to get the same, but it's showing cityname partly when cityname has space in between. Works fine when cityname is a single word.

The current query which you gave is showing part of the cityname when cityname has space in between( ex: Plain City , Mason city, but query is returning only Plain , Mason.
It's skipping "City" word from the cityname. I will post source data and output data that I need in the next reply. For some reason there is a chars limitation while posting.

Thanks,
RH
SQL9 246 Reputation points

2021-06-01T15:23:36.467+00:00

Address could have cityname with "City", "Ville" or anything. I need full cityname like below output format.

My source data looks like below: I am using 2017 SQL version.
declare @Address table (FullAddress varchar(150))
insert into @Address (FullAddress)
Select '5500 North AVENUE, APT. 116 PLAIN CITY OH 60053'
union all
Select '12536 S INDIANA AVE CHICAGO IL 60625'
union all
Select 'W MAIN ST PO BOX 111 ADDISON VA 24293'
union all
select '179 ARMY TRAIL RD BLOOMINGDALE IL 60108'
union all
select '100 NORTH MAIN MASON CITY IL 626641104'
union all
select '123 West Street EDISON NJ 34561'

Output should look like in below format:

PLAIN CITY OH 60053
CHICAGO IL 60625
ADDISON VA 24293
BLOOMINGDALE IL 60108
MASON CITY IL 626641104
EDISON NJ 34561

Thanks,
RH
Guoxiong 8,221 Reputation points

2021-06-01T16:18:36.73+00:00

Can all of cities in the Address table be found in the City table?
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MelissaMa-msft 24,246 Reputation points Moderator

2021-06-03T00:59:38.757+00:00

Hi @SQL9 ,

Thanks for your update.

Please help refer the latest answers and check whether any of them is working.

Please remember to accept the answers if they helped. Your action would be helpful to other users who encounter the same issue and read this thread.

Thank you for understanding!

Best regards,
Melissa

Answer accepted by question author

MelissaMa-msft 24,246 Moderator

Hi @SQL9 ,

Thanks for your update.

If you could list all cityname with "City", "Ville" or anything, you could refer below:

;with cte as (  
select  ROW_NUMBER() Over (order by (select null)) addressid,FullAddress from @Address)  
,cte1 as (  
SELECT ROW_NUMBER() Over (partition by addressid order by (select null) ) id, value,addressid FROM cte  
cross apply STRING_SPLIT(FullAddress, SPACE(1)))  
,cte2 as (  
select ROW_NUMBER() Over (partition by addressid order by id desc) rn ,value ,addressid  
from cte1)  
,cte3 as (  
select a.*,b.value1   
from cte2 a  
left join (select addressid,value value1 from cte2 where rn=3) b on a.addressid=b.addressid)  
select STRING_AGG(value,SPACE(1)) WITHIN GROUP (ORDER BY rn DESC) Result  
from cte3  
where value1 in ('city','Ville') and rn<=4  
group by addressid  
union   
select STRING_AGG(value,SPACE(1)) WITHIN GROUP (ORDER BY rn DESC) Result  
from cte3  
where value1 not in ('city','Ville') and rn<4  
group by addressid

Output:

Result  
ADDISON VA 24293  
BLOOMINGDALE IL 60108  
CHICAGO IL 60625  
EDISON NJ 34561  
MASON CITY IL 626641104  
PLAIN CITY OH 60053

Or you could refer another method using user-defined function.

Firstly create one function as below:

CREATE FUNCTION dbo.GetSplitString  
(  
   @List       VARCHAR(MAX),  
   @Delimiter  VARCHAR(255),  
   @ElementNumber int  
)  
RETURNS VARCHAR(4000)  
AS  
BEGIN  
   DECLARE @result varchar(4000)      
   DECLARE @Items TABLE ( position int IDENTITY PRIMARY KEY,  
                          Item VARCHAR(4000)  
                         )    
  
   DECLARE @ll INT = LEN(@List) + 1, @ld INT = LEN(@Delimiter);    
  
   WITH a AS  
   (  
       SELECT  
           [start] = 1,  
           [end]   = COALESCE(NULLIF(CHARINDEX(@Delimiter,   
                       @List, @ld), 0), @ll),  
           [value] = SUBSTRING(@List, 1,   
                     COALESCE(NULLIF(CHARINDEX(@Delimiter,   
                       @List, @ld), 0), @ll) - 1)  
       UNION ALL  
       SELECT  
           [start] = CONVERT(INT, [end]) + @ld,  
           [end]   = COALESCE(NULLIF(CHARINDEX(@Delimiter,   
                       @List, [end] + @ld), 0), @ll),  
           [value] = SUBSTRING(@List, [end] + @ld,   
                     COALESCE(NULLIF(CHARINDEX(@Delimiter,   
                       @List, [end] + @ld), 0), @ll)-[end]-@ld)  
       FROM a  
       WHERE [end] < @ll  
   )  
   INSERT @Items SELECT [value]  
   FROM a  
   WHERE LEN([value]) > 0  
   OPTION (MAXRECURSION 0);  
  
   SELECT @result=Item  
   FROM @Items  
   WHERE position=@ElementNumber  
     
   RETURN @result;  
END  
GO

Then call this function as below:

SELECT case when REVERSE(dbo.GetSplitString(REVERSE(replace(FullAddress,' ',',')),',',3)) in ('city','Ville')  
then right(FullAddress,charindex(' ', REVERSE(FullAddress),charindex(' ', REVERSE(FullAddress),charindex(' ', REVERSE(FullAddress), (charindex(' ', REVERSE(FullAddress), 1))+1)+1)+1))  
else right(FullAddress,charindex(' ', REVERSE(FullAddress),charindex(' ', REVERSE(FullAddress), (charindex(' ', REVERSE(FullAddress), 1))+1)+1))  
end result  
from @Address

Best regards,
Melissa

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SQL9 246 Reputation points

2021-06-09T17:06:58.153+00:00

Thank you Melissa, The options which you have given worked as expected. Sorry for the delay response. I appreciate your help.

Thank you,
RH

5 additional answers

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