Find records with same string with extra character
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3
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OK, so I have a Microsoft SQL Server 2014 database table called owner
with around 90,000 records with owner information, another called vehicle
with vehicle information
Owner_Name owner_id V_name owner_id exempt
------------------------------------- ------------------------------
JACOB JAMISON & JESSICA 35 Civic 35 H3
JACOB JAMISON M & JESSICA B 39 Accord 39 H3
BLACKSON BARRINGTON 56 Bugatti 56 H6
BLACKSON BARRINGTON H 98 SSC 98 H7
BRUSTER MICHAEL 107 Corvette 107 H9
I'm trying to find all records that have more than one exemption on a vehicle
( H0
means no exemption). This code below has worked well, as long as the name is exactly the same. However, if there's a variation, such as an extra letter or if it's entered backwards, then it won't return those records. I've looked at things like SOUNDEX
, but this wouldn't work in my scenario.
SELECT Owner_name
, COUNT(Owner_name) AS 'xNameAppears'
, COUNT(v.exempt) AS 'ExemptionCount'
FROM owner o
INNER JOIN vehicle V ON V.owner_id = o.owner_id
WHERE v.exempt <> 'H0'
GROUP BY O.owner_name
HAVING COUNT(v.exempt) > 1
Is there a solution that would allow me to return records like so, not knowing which owner_name
's may be similar? Basically trying to get the server to search through the owner_name
column and if there's a similarity such as
JACOB JAMISON & JESSICA
and JACOB JAMISON M & JESSICA B
then it'll return those records like so:
Owner_Name xNameAppears ExemptCount
-------------------------------------------------------------
JACOB JAMISON & JESSICA 2 2
JACOB JAMISON M & JESSICA B 2 2
BLACKSON BARRINGTON 2 2
BLACKSON BARRINGTON H 2 2
Thank you in advance!
sql-server sql-server-2014 full-text-search string-searching
New contributor
add a comment |
up vote
3
down vote
favorite
OK, so I have a Microsoft SQL Server 2014 database table called owner
with around 90,000 records with owner information, another called vehicle
with vehicle information
Owner_Name owner_id V_name owner_id exempt
------------------------------------- ------------------------------
JACOB JAMISON & JESSICA 35 Civic 35 H3
JACOB JAMISON M & JESSICA B 39 Accord 39 H3
BLACKSON BARRINGTON 56 Bugatti 56 H6
BLACKSON BARRINGTON H 98 SSC 98 H7
BRUSTER MICHAEL 107 Corvette 107 H9
I'm trying to find all records that have more than one exemption on a vehicle
( H0
means no exemption). This code below has worked well, as long as the name is exactly the same. However, if there's a variation, such as an extra letter or if it's entered backwards, then it won't return those records. I've looked at things like SOUNDEX
, but this wouldn't work in my scenario.
SELECT Owner_name
, COUNT(Owner_name) AS 'xNameAppears'
, COUNT(v.exempt) AS 'ExemptionCount'
FROM owner o
INNER JOIN vehicle V ON V.owner_id = o.owner_id
WHERE v.exempt <> 'H0'
GROUP BY O.owner_name
HAVING COUNT(v.exempt) > 1
Is there a solution that would allow me to return records like so, not knowing which owner_name
's may be similar? Basically trying to get the server to search through the owner_name
column and if there's a similarity such as
JACOB JAMISON & JESSICA
and JACOB JAMISON M & JESSICA B
then it'll return those records like so:
Owner_Name xNameAppears ExemptCount
-------------------------------------------------------------
JACOB JAMISON & JESSICA 2 2
JACOB JAMISON M & JESSICA B 2 2
BLACKSON BARRINGTON 2 2
BLACKSON BARRINGTON H 2 2
Thank you in advance!
sql-server sql-server-2014 full-text-search string-searching
New contributor
1
SureSOUNDEX
is suitable? The following all return the same valueSOUNDEX('JACOB JAMISON & JESSICA '), SOUNDEX('Jacob Zuma'), SOUNDEX('Jacob Willekens'), SOUNDEX('Jacob S. Coxey, Sr.'), SOUNDEX('Juegief zzzzzzzzzzzz')
– Martin Smith
11 hours ago
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
OK, so I have a Microsoft SQL Server 2014 database table called owner
with around 90,000 records with owner information, another called vehicle
with vehicle information
Owner_Name owner_id V_name owner_id exempt
------------------------------------- ------------------------------
JACOB JAMISON & JESSICA 35 Civic 35 H3
JACOB JAMISON M & JESSICA B 39 Accord 39 H3
BLACKSON BARRINGTON 56 Bugatti 56 H6
BLACKSON BARRINGTON H 98 SSC 98 H7
BRUSTER MICHAEL 107 Corvette 107 H9
I'm trying to find all records that have more than one exemption on a vehicle
( H0
means no exemption). This code below has worked well, as long as the name is exactly the same. However, if there's a variation, such as an extra letter or if it's entered backwards, then it won't return those records. I've looked at things like SOUNDEX
, but this wouldn't work in my scenario.
SELECT Owner_name
, COUNT(Owner_name) AS 'xNameAppears'
, COUNT(v.exempt) AS 'ExemptionCount'
FROM owner o
INNER JOIN vehicle V ON V.owner_id = o.owner_id
WHERE v.exempt <> 'H0'
GROUP BY O.owner_name
HAVING COUNT(v.exempt) > 1
Is there a solution that would allow me to return records like so, not knowing which owner_name
's may be similar? Basically trying to get the server to search through the owner_name
column and if there's a similarity such as
JACOB JAMISON & JESSICA
and JACOB JAMISON M & JESSICA B
then it'll return those records like so:
Owner_Name xNameAppears ExemptCount
-------------------------------------------------------------
JACOB JAMISON & JESSICA 2 2
JACOB JAMISON M & JESSICA B 2 2
BLACKSON BARRINGTON 2 2
BLACKSON BARRINGTON H 2 2
Thank you in advance!
sql-server sql-server-2014 full-text-search string-searching
New contributor
OK, so I have a Microsoft SQL Server 2014 database table called owner
with around 90,000 records with owner information, another called vehicle
with vehicle information
Owner_Name owner_id V_name owner_id exempt
------------------------------------- ------------------------------
JACOB JAMISON & JESSICA 35 Civic 35 H3
JACOB JAMISON M & JESSICA B 39 Accord 39 H3
BLACKSON BARRINGTON 56 Bugatti 56 H6
BLACKSON BARRINGTON H 98 SSC 98 H7
BRUSTER MICHAEL 107 Corvette 107 H9
I'm trying to find all records that have more than one exemption on a vehicle
( H0
means no exemption). This code below has worked well, as long as the name is exactly the same. However, if there's a variation, such as an extra letter or if it's entered backwards, then it won't return those records. I've looked at things like SOUNDEX
, but this wouldn't work in my scenario.
SELECT Owner_name
, COUNT(Owner_name) AS 'xNameAppears'
, COUNT(v.exempt) AS 'ExemptionCount'
FROM owner o
INNER JOIN vehicle V ON V.owner_id = o.owner_id
WHERE v.exempt <> 'H0'
GROUP BY O.owner_name
HAVING COUNT(v.exempt) > 1
Is there a solution that would allow me to return records like so, not knowing which owner_name
's may be similar? Basically trying to get the server to search through the owner_name
column and if there's a similarity such as
JACOB JAMISON & JESSICA
and JACOB JAMISON M & JESSICA B
then it'll return those records like so:
Owner_Name xNameAppears ExemptCount
-------------------------------------------------------------
JACOB JAMISON & JESSICA 2 2
JACOB JAMISON M & JESSICA B 2 2
BLACKSON BARRINGTON 2 2
BLACKSON BARRINGTON H 2 2
Thank you in advance!
sql-server sql-server-2014 full-text-search string-searching
sql-server sql-server-2014 full-text-search string-searching
New contributor
New contributor
New contributor
asked 14 hours ago
MindXpert
183
183
New contributor
New contributor
1
SureSOUNDEX
is suitable? The following all return the same valueSOUNDEX('JACOB JAMISON & JESSICA '), SOUNDEX('Jacob Zuma'), SOUNDEX('Jacob Willekens'), SOUNDEX('Jacob S. Coxey, Sr.'), SOUNDEX('Juegief zzzzzzzzzzzz')
– Martin Smith
11 hours ago
add a comment |
1
SureSOUNDEX
is suitable? The following all return the same valueSOUNDEX('JACOB JAMISON & JESSICA '), SOUNDEX('Jacob Zuma'), SOUNDEX('Jacob Willekens'), SOUNDEX('Jacob S. Coxey, Sr.'), SOUNDEX('Juegief zzzzzzzzzzzz')
– Martin Smith
11 hours ago
1
1
Sure
SOUNDEX
is suitable? The following all return the same value SOUNDEX('JACOB JAMISON & JESSICA '), SOUNDEX('Jacob Zuma'), SOUNDEX('Jacob Willekens'), SOUNDEX('Jacob S. Coxey, Sr.'), SOUNDEX('Juegief zzzzzzzzzzzz')
– Martin Smith
11 hours ago
Sure
SOUNDEX
is suitable? The following all return the same value SOUNDEX('JACOB JAMISON & JESSICA '), SOUNDEX('Jacob Zuma'), SOUNDEX('Jacob Willekens'), SOUNDEX('Jacob S. Coxey, Sr.'), SOUNDEX('Juegief zzzzzzzzzzzz')
– Martin Smith
11 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
The SOUNDEX function can be applied to a column as well.
But since
there's thousands like that
I wouldn't suggest just writing a query to join on a function to do that.
This will likely not perform very well on larger tables:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON SOUNDEX(v2.Owner_Name) = SOUNDEX(v.Owner_Name)
AND v2.Owner_Name <> v.Owner_Name;
I'd rather do something that will make finding this easier in the long-term.
Here's an example:
CREATE TABLE dbo.vehicle (Owner_Name VARCHAR(50));
INSERT dbo.vehicle ( Owner_Name )
SELECT *
FROM (
VALUES
('JACOB JAMISON & JESSICA'),
('JACOB JAMISON M & JESSICA B'),
('BLACKSON BARRINGTON'),
('BLACKSON BARRINGTON H'),
('BRUSTER MICHAEL')
) AS x (Owner_Name);
I'm going to add a computed column based on the function, and then add an index to aid my query.
ALTER TABLE dbo.vehicle ADD Owner_Soundex AS SOUNDEX(Owner_Name);
CREATE INDEX ix_whatever ON dbo.vehicle (Owner_Soundex, Owner_Name);
Validate that everything looks good...
SELECT *
FROM dbo.vehicle AS v
Use a query like this to find imprecise matches:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON v2.Owner_Soundex = v.Owner_Soundex
AND v2.Owner_Name <> v.Owner_Name;
You sir, are the man. I didn't think to useSOUNDEX
in that manner. This works very well! Indeed that second way is more efficient performance wise. Thank you.
– MindXpert
12 hours ago
@user1227080 happy to help!
– sp_BlitzErik
12 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
The SOUNDEX function can be applied to a column as well.
But since
there's thousands like that
I wouldn't suggest just writing a query to join on a function to do that.
This will likely not perform very well on larger tables:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON SOUNDEX(v2.Owner_Name) = SOUNDEX(v.Owner_Name)
AND v2.Owner_Name <> v.Owner_Name;
I'd rather do something that will make finding this easier in the long-term.
Here's an example:
CREATE TABLE dbo.vehicle (Owner_Name VARCHAR(50));
INSERT dbo.vehicle ( Owner_Name )
SELECT *
FROM (
VALUES
('JACOB JAMISON & JESSICA'),
('JACOB JAMISON M & JESSICA B'),
('BLACKSON BARRINGTON'),
('BLACKSON BARRINGTON H'),
('BRUSTER MICHAEL')
) AS x (Owner_Name);
I'm going to add a computed column based on the function, and then add an index to aid my query.
ALTER TABLE dbo.vehicle ADD Owner_Soundex AS SOUNDEX(Owner_Name);
CREATE INDEX ix_whatever ON dbo.vehicle (Owner_Soundex, Owner_Name);
Validate that everything looks good...
SELECT *
FROM dbo.vehicle AS v
Use a query like this to find imprecise matches:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON v2.Owner_Soundex = v.Owner_Soundex
AND v2.Owner_Name <> v.Owner_Name;
You sir, are the man. I didn't think to useSOUNDEX
in that manner. This works very well! Indeed that second way is more efficient performance wise. Thank you.
– MindXpert
12 hours ago
@user1227080 happy to help!
– sp_BlitzErik
12 hours ago
add a comment |
up vote
5
down vote
accepted
The SOUNDEX function can be applied to a column as well.
But since
there's thousands like that
I wouldn't suggest just writing a query to join on a function to do that.
This will likely not perform very well on larger tables:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON SOUNDEX(v2.Owner_Name) = SOUNDEX(v.Owner_Name)
AND v2.Owner_Name <> v.Owner_Name;
I'd rather do something that will make finding this easier in the long-term.
Here's an example:
CREATE TABLE dbo.vehicle (Owner_Name VARCHAR(50));
INSERT dbo.vehicle ( Owner_Name )
SELECT *
FROM (
VALUES
('JACOB JAMISON & JESSICA'),
('JACOB JAMISON M & JESSICA B'),
('BLACKSON BARRINGTON'),
('BLACKSON BARRINGTON H'),
('BRUSTER MICHAEL')
) AS x (Owner_Name);
I'm going to add a computed column based on the function, and then add an index to aid my query.
ALTER TABLE dbo.vehicle ADD Owner_Soundex AS SOUNDEX(Owner_Name);
CREATE INDEX ix_whatever ON dbo.vehicle (Owner_Soundex, Owner_Name);
Validate that everything looks good...
SELECT *
FROM dbo.vehicle AS v
Use a query like this to find imprecise matches:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON v2.Owner_Soundex = v.Owner_Soundex
AND v2.Owner_Name <> v.Owner_Name;
You sir, are the man. I didn't think to useSOUNDEX
in that manner. This works very well! Indeed that second way is more efficient performance wise. Thank you.
– MindXpert
12 hours ago
@user1227080 happy to help!
– sp_BlitzErik
12 hours ago
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
The SOUNDEX function can be applied to a column as well.
But since
there's thousands like that
I wouldn't suggest just writing a query to join on a function to do that.
This will likely not perform very well on larger tables:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON SOUNDEX(v2.Owner_Name) = SOUNDEX(v.Owner_Name)
AND v2.Owner_Name <> v.Owner_Name;
I'd rather do something that will make finding this easier in the long-term.
Here's an example:
CREATE TABLE dbo.vehicle (Owner_Name VARCHAR(50));
INSERT dbo.vehicle ( Owner_Name )
SELECT *
FROM (
VALUES
('JACOB JAMISON & JESSICA'),
('JACOB JAMISON M & JESSICA B'),
('BLACKSON BARRINGTON'),
('BLACKSON BARRINGTON H'),
('BRUSTER MICHAEL')
) AS x (Owner_Name);
I'm going to add a computed column based on the function, and then add an index to aid my query.
ALTER TABLE dbo.vehicle ADD Owner_Soundex AS SOUNDEX(Owner_Name);
CREATE INDEX ix_whatever ON dbo.vehicle (Owner_Soundex, Owner_Name);
Validate that everything looks good...
SELECT *
FROM dbo.vehicle AS v
Use a query like this to find imprecise matches:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON v2.Owner_Soundex = v.Owner_Soundex
AND v2.Owner_Name <> v.Owner_Name;
The SOUNDEX function can be applied to a column as well.
But since
there's thousands like that
I wouldn't suggest just writing a query to join on a function to do that.
This will likely not perform very well on larger tables:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON SOUNDEX(v2.Owner_Name) = SOUNDEX(v.Owner_Name)
AND v2.Owner_Name <> v.Owner_Name;
I'd rather do something that will make finding this easier in the long-term.
Here's an example:
CREATE TABLE dbo.vehicle (Owner_Name VARCHAR(50));
INSERT dbo.vehicle ( Owner_Name )
SELECT *
FROM (
VALUES
('JACOB JAMISON & JESSICA'),
('JACOB JAMISON M & JESSICA B'),
('BLACKSON BARRINGTON'),
('BLACKSON BARRINGTON H'),
('BRUSTER MICHAEL')
) AS x (Owner_Name);
I'm going to add a computed column based on the function, and then add an index to aid my query.
ALTER TABLE dbo.vehicle ADD Owner_Soundex AS SOUNDEX(Owner_Name);
CREATE INDEX ix_whatever ON dbo.vehicle (Owner_Soundex, Owner_Name);
Validate that everything looks good...
SELECT *
FROM dbo.vehicle AS v
Use a query like this to find imprecise matches:
SELECT *
FROM dbo.vehicle AS v
JOIN dbo.vehicle AS v2
ON v2.Owner_Soundex = v.Owner_Soundex
AND v2.Owner_Name <> v.Owner_Name;
answered 13 hours ago
sp_BlitzErik
20.6k1262102
20.6k1262102
You sir, are the man. I didn't think to useSOUNDEX
in that manner. This works very well! Indeed that second way is more efficient performance wise. Thank you.
– MindXpert
12 hours ago
@user1227080 happy to help!
– sp_BlitzErik
12 hours ago
add a comment |
You sir, are the man. I didn't think to useSOUNDEX
in that manner. This works very well! Indeed that second way is more efficient performance wise. Thank you.
– MindXpert
12 hours ago
@user1227080 happy to help!
– sp_BlitzErik
12 hours ago
You sir, are the man. I didn't think to use
SOUNDEX
in that manner. This works very well! Indeed that second way is more efficient performance wise. Thank you.– MindXpert
12 hours ago
You sir, are the man. I didn't think to use
SOUNDEX
in that manner. This works very well! Indeed that second way is more efficient performance wise. Thank you.– MindXpert
12 hours ago
@user1227080 happy to help!
– sp_BlitzErik
12 hours ago
@user1227080 happy to help!
– sp_BlitzErik
12 hours ago
add a comment |
MindXpert is a new contributor. Be nice, and check out our Code of Conduct.
MindXpert is a new contributor. Be nice, and check out our Code of Conduct.
MindXpert is a new contributor. Be nice, and check out our Code of Conduct.
MindXpert is a new contributor. Be nice, and check out our Code of Conduct.
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1
Sure
SOUNDEX
is suitable? The following all return the same valueSOUNDEX('JACOB JAMISON & JESSICA '), SOUNDEX('Jacob Zuma'), SOUNDEX('Jacob Willekens'), SOUNDEX('Jacob S. Coxey, Sr.'), SOUNDEX('Juegief zzzzzzzzzzzz')
– Martin Smith
11 hours ago