ms-access 2010: count duplicate names per household address - sql

Question

I am currently working with a spreadsheet in MS Access 2010 which contains about 130k rows of information about people who voted in a local election recently. Each row has their residential information (street name, number, postcode etc.) and personal information (title, surname, forename, middle name, DOB etc.). Each row represents an individual person rather than a household (therefore in many cases the same residential address appears more than once as more than one person resides in a particular household).
What I want to achieve is basically to create a new field in this dataset called 'count'. I want this field to give me a count of how many different surnames reside at a single address.
Is there an SQL script that will allow me to do this in Access 2010?
+------------------+----------+-------+---------+----------+-------------+
| PROPERTYADDRESS1 | POSTCODE | TITLE | SURNAME | FORENAME | MIDDLE_NAME |
+------------------+----------+-------+---------+----------+-------------+
FAKEADDRESS1 EEE 5GG MR BLOGGS JOE N
FAKEADDRESS2 EEE 5BB MRS BLOGGS SUZANNE P
FAKEADDRESS3 EEE 5RG MS SMITH PAULINE S
FAKEADDRESS4 EEE 4BV DR JONES ANNE D
FAKEADDRESS5 EEE 3AS MR TAYLOR STUART A
The following syntax has got me close so far:
SELECT COUNT(electoral.SURNAME)
FROM electoral
GROUP BY electoral.UPRN
However, instead of returning me all 130k odd rows, it only returns me around 67k rows. Is there anything I can do to the syntax to achieve the same result, but just returning every single row?
Any help is greatly appreciated!
Thanks

Answer 1

You could use something like this:
select *,
count(surname) over (partition by householdName)
from myTable
If you have only one column which contains the name,
ex: Rob Adams
then you can do this to have all the surnames in a different column so it will be easier in the select:
SELECT LEFT('HELLO WORLD',CHARINDEX(' ','HELLO WORLD')-1)
in our example:
select right (surmane, charindex (' ',surname)-1) as surname
example on how to use charindex, left and right here:
http://social.technet.microsoft.com/wiki/contents/articles/17948.t-sql-right-left-substring-and-charindex-functions.aspx
if there are any questions, leave a comment.
EDIT: I edited the query, had a syntax error, please try it again. This works on sql server.
here is an example:
create table #temp (id int, PropertyAddress varchar(50), surname varchar(50), forname varchar(50))
insert into #temp values
(1, 'hiddenBase', 'Adamns' , 'Kara' ),
(2, 'hiddenBase', 'Adamns' , 'Anne' ),
(3, 'hiddenBase', 'Adamns' , 'John' ),
(4, 'QueensResidence', 'Queen' , 'Oliver' ),
(5, 'QueensResidence', 'Queen' , 'Moira' ),
(6, 'superSecretBase', 'Diggle' , 'John' ),
(7, 'NandaParbat', 'Merlin' , 'Malcom' )
select * from #temp
select *,
count (surname) over (partition by PropertyAddress) as CountMembers
from #temp
gives:
1 hiddenBase Adamns Kara 3
2 hiddenBase Adamns Anne 3
3 hiddenBase Adamns John 3
7 NandaParbat Merlin Malcom 1
4 QueensResidence Queen Oliver 2
5 QueensResidence Queen Moira 2
6 superSecretBase Diggle John 1
Your query should look like this:
select *,
count (SURNAME) over (partition by PropertyAddress) as CountFamilyMembers
from electoral
EDIT
If over partition by isn't supported, then I guess you can get to your desired result by using group by
select *,
count (SURNAME) over (partition by PropertyAddress) as CountFamilyMembers
from electoral
group by -- put here the fields in the select (one by one), however you can't write group by *

Answer 2

GROUP BY creates an aggregate query, so it's by design that you get fewer records (one per UPRN).
To get the count for each row in the original table, you can join the table with the aggregate query:
SELECT electoral.*, elCount.NumberOfPeople
FROM electoral
INNER JOIN
(
SELECT UPRN, COUNT(*) AS NumberOfPeople
FROM electoral
GROUP BY UPRN
) AS elCount
ON electoral.UPRN = elCount.UPRN

Answer 3

Given the update I want to post another answer. Try it like this:
create table #temp2 ( PropertyAddress1 varchar(50), POSTCODE varchar(20), TITLE varchar (20),
surname varchar(50), FORENAME varchar(50), MIDDLE_NAME varchar (50) )
insert into #temp2 values
('FAKEADDRESS1', 'EEE 5GG', 'MR', 'BLOGGS', 'JOE', 'N'),
('FAKEADDRESS1', 'EEE 5BB', 'MRS', 'BLOGGS', 'SUZANNE', 'P'),
('FAKEADDRESS2', 'EEE 5RG', 'MS', 'SMITH', 'PAULINE', 'S'),
('FAKEADDRESS3', 'EEE 4BV', 'DR', 'JONES', 'ANNE', 'D'),
('FAKEADDRESS4', 'EEE 3AS', 'MR', 'TAYLOR', 'STUART', 'A')
select PropertyAddress1, surname,count (#temp2.surname) as CountADD
into #countTemp
from #temp2
group by PropertyAddress1, surname
select * from #temp2 t2
left join #countTemp ct
on t2.PropertyAddress1 = ct.PropertyAddress1 and t2.surname = ct.surname
This yields:
PropertyAddress1 POSTCODE TITLE surname FORENAME MIDDLE_NAME PropertyAddress1 surname CountADD
FAKEADDRESS1 EEE 5GG MR BLOGGS JOE N FAKEADDRESS1 BLOGGS 2
FAKEADDRESS1 EEE 5BB MRS BLOGGS SUZANNE P FAKEADDRESS1 BLOGGS 2
FAKEADDRESS2 EEE 5RG MS SMITH PAULINE S FAKEADDRESS2 SMITH 1
FAKEADDRESS3 EEE 4BV DR JONES ANNE D FAKEADDRESS3 JONES 1
FAKEADDRESS4 EEE 3AS MR TAYLOR STUART A FAKEADDRESS4 TAYLOR 1