Query to get subjects of interest for all User Y where Y shares >=3 interests with a User X
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These are two tables from a part of supposed Twitter like database where users can follow other users. The User.name field is unique.
mysql> select uID, name from User;
+-----+-------------------+
| uID | name |
+-----+-------------------+
| 1 | Alice |
| 2 | Bob |
| 5 | Iron Maiden |
| 4 | Judas Priest |
| 6 | Lesser Known Band |
| 3 | Metallica |
+-----+-------------------+
6 rows in set (0.00 sec)
mysql> select * from Follower;
+-----------+------------+
| subjectID | observerID |
+-----------+------------+
| 3 | 1 |
| 4 | 1 |
| 5 | 1 |
| 6 | 1 |
| 3 | 2 |
| 4 | 2 |
| 5 | 2 |
+-----------+------------+
7 rows in set (0.00 sec)
mysql> call newFollowSuggestionsForName('Bob');
+-------------------+
| name |
+-------------------+
| Lesser Known Band |
+-------------------+
1 row in set (0.00 sec)
I want to make an operation that will suggest for a user X a list of users they may be interested in following. I thought one heuristic could be to show X for all y who user y follows where X and y follow at least 3 of the same Users. Below is the SQL I came up with to do this. My question is if it could be done more efficiently or nicer in some other ways.
DELIMITER //
CREATE PROCEDURE newFollowSuggestionsForName(IN in_name CHAR(60))
BEGIN
DECLARE xuid INT;
SET xuid = (select uID from User where name=in_name);
select name
from User, (select subjectID
from follower
where observerID in (
select observerID
from Follower
where observerID<>xuid and subjectID in (select subjectID from Follower where observerID=xuid)
group by observerID
having count(*)>=3
)
) as T
where uID = T.subjectID and not exists (select * from Follower where subjectID=T.subjectID and observerID=xuid);
END //
DELIMITER ;
mysql sql
add a comment |
up vote
1
down vote
favorite
These are two tables from a part of supposed Twitter like database where users can follow other users. The User.name field is unique.
mysql> select uID, name from User;
+-----+-------------------+
| uID | name |
+-----+-------------------+
| 1 | Alice |
| 2 | Bob |
| 5 | Iron Maiden |
| 4 | Judas Priest |
| 6 | Lesser Known Band |
| 3 | Metallica |
+-----+-------------------+
6 rows in set (0.00 sec)
mysql> select * from Follower;
+-----------+------------+
| subjectID | observerID |
+-----------+------------+
| 3 | 1 |
| 4 | 1 |
| 5 | 1 |
| 6 | 1 |
| 3 | 2 |
| 4 | 2 |
| 5 | 2 |
+-----------+------------+
7 rows in set (0.00 sec)
mysql> call newFollowSuggestionsForName('Bob');
+-------------------+
| name |
+-------------------+
| Lesser Known Band |
+-------------------+
1 row in set (0.00 sec)
I want to make an operation that will suggest for a user X a list of users they may be interested in following. I thought one heuristic could be to show X for all y who user y follows where X and y follow at least 3 of the same Users. Below is the SQL I came up with to do this. My question is if it could be done more efficiently or nicer in some other ways.
DELIMITER //
CREATE PROCEDURE newFollowSuggestionsForName(IN in_name CHAR(60))
BEGIN
DECLARE xuid INT;
SET xuid = (select uID from User where name=in_name);
select name
from User, (select subjectID
from follower
where observerID in (
select observerID
from Follower
where observerID<>xuid and subjectID in (select subjectID from Follower where observerID=xuid)
group by observerID
having count(*)>=3
)
) as T
where uID = T.subjectID and not exists (select * from Follower where subjectID=T.subjectID and observerID=xuid);
END //
DELIMITER ;
mysql sql
Why do you inputin_name
as Char. You must be having its integer ID value also. My suggestion is to use that instead. It will be more performant. Moreoever, what happens if there are two users with same name(s) ?
– Madhur Bhaiya
Nov 11 at 8:22
Also, do you have access to latest version of MySQL (version 8.0.2 and above) ?
– Madhur Bhaiya
Nov 11 at 8:26
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
These are two tables from a part of supposed Twitter like database where users can follow other users. The User.name field is unique.
mysql> select uID, name from User;
+-----+-------------------+
| uID | name |
+-----+-------------------+
| 1 | Alice |
| 2 | Bob |
| 5 | Iron Maiden |
| 4 | Judas Priest |
| 6 | Lesser Known Band |
| 3 | Metallica |
+-----+-------------------+
6 rows in set (0.00 sec)
mysql> select * from Follower;
+-----------+------------+
| subjectID | observerID |
+-----------+------------+
| 3 | 1 |
| 4 | 1 |
| 5 | 1 |
| 6 | 1 |
| 3 | 2 |
| 4 | 2 |
| 5 | 2 |
+-----------+------------+
7 rows in set (0.00 sec)
mysql> call newFollowSuggestionsForName('Bob');
+-------------------+
| name |
+-------------------+
| Lesser Known Band |
+-------------------+
1 row in set (0.00 sec)
I want to make an operation that will suggest for a user X a list of users they may be interested in following. I thought one heuristic could be to show X for all y who user y follows where X and y follow at least 3 of the same Users. Below is the SQL I came up with to do this. My question is if it could be done more efficiently or nicer in some other ways.
DELIMITER //
CREATE PROCEDURE newFollowSuggestionsForName(IN in_name CHAR(60))
BEGIN
DECLARE xuid INT;
SET xuid = (select uID from User where name=in_name);
select name
from User, (select subjectID
from follower
where observerID in (
select observerID
from Follower
where observerID<>xuid and subjectID in (select subjectID from Follower where observerID=xuid)
group by observerID
having count(*)>=3
)
) as T
where uID = T.subjectID and not exists (select * from Follower where subjectID=T.subjectID and observerID=xuid);
END //
DELIMITER ;
mysql sql
These are two tables from a part of supposed Twitter like database where users can follow other users. The User.name field is unique.
mysql> select uID, name from User;
+-----+-------------------+
| uID | name |
+-----+-------------------+
| 1 | Alice |
| 2 | Bob |
| 5 | Iron Maiden |
| 4 | Judas Priest |
| 6 | Lesser Known Band |
| 3 | Metallica |
+-----+-------------------+
6 rows in set (0.00 sec)
mysql> select * from Follower;
+-----------+------------+
| subjectID | observerID |
+-----------+------------+
| 3 | 1 |
| 4 | 1 |
| 5 | 1 |
| 6 | 1 |
| 3 | 2 |
| 4 | 2 |
| 5 | 2 |
+-----------+------------+
7 rows in set (0.00 sec)
mysql> call newFollowSuggestionsForName('Bob');
+-------------------+
| name |
+-------------------+
| Lesser Known Band |
+-------------------+
1 row in set (0.00 sec)
I want to make an operation that will suggest for a user X a list of users they may be interested in following. I thought one heuristic could be to show X for all y who user y follows where X and y follow at least 3 of the same Users. Below is the SQL I came up with to do this. My question is if it could be done more efficiently or nicer in some other ways.
DELIMITER //
CREATE PROCEDURE newFollowSuggestionsForName(IN in_name CHAR(60))
BEGIN
DECLARE xuid INT;
SET xuid = (select uID from User where name=in_name);
select name
from User, (select subjectID
from follower
where observerID in (
select observerID
from Follower
where observerID<>xuid and subjectID in (select subjectID from Follower where observerID=xuid)
group by observerID
having count(*)>=3
)
) as T
where uID = T.subjectID and not exists (select * from Follower where subjectID=T.subjectID and observerID=xuid);
END //
DELIMITER ;
mysql sql
mysql sql
asked Nov 10 at 23:52
bagel_lord
29113
29113
Why do you inputin_name
as Char. You must be having its integer ID value also. My suggestion is to use that instead. It will be more performant. Moreoever, what happens if there are two users with same name(s) ?
– Madhur Bhaiya
Nov 11 at 8:22
Also, do you have access to latest version of MySQL (version 8.0.2 and above) ?
– Madhur Bhaiya
Nov 11 at 8:26
add a comment |
Why do you inputin_name
as Char. You must be having its integer ID value also. My suggestion is to use that instead. It will be more performant. Moreoever, what happens if there are two users with same name(s) ?
– Madhur Bhaiya
Nov 11 at 8:22
Also, do you have access to latest version of MySQL (version 8.0.2 and above) ?
– Madhur Bhaiya
Nov 11 at 8:26
Why do you input
in_name
as Char. You must be having its integer ID value also. My suggestion is to use that instead. It will be more performant. Moreoever, what happens if there are two users with same name(s) ?– Madhur Bhaiya
Nov 11 at 8:22
Why do you input
in_name
as Char. You must be having its integer ID value also. My suggestion is to use that instead. It will be more performant. Moreoever, what happens if there are two users with same name(s) ?– Madhur Bhaiya
Nov 11 at 8:22
Also, do you have access to latest version of MySQL (version 8.0.2 and above) ?
– Madhur Bhaiya
Nov 11 at 8:26
Also, do you have access to latest version of MySQL (version 8.0.2 and above) ?
– Madhur Bhaiya
Nov 11 at 8:26
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
Consider the following refactored SQL code (untested without data) for use in stored procedure.
select u.`name`
from `User` u
inner join
(select subf.observerID, subf.subjectID
from follower subf
where subf.observerID <> xuid
) f
on u.UID = f.subjectID
inner join
(select f1.observerID
from follower f1
inner join follower f2
on f1.subjectID = f2.subjectID
and f1.observerID <> xuid
and f2.observerID = xuid
group by f1.observerID
having count(*) >= 3
) o
on f.observerID = o.observerID
add a comment |
up vote
0
down vote
I think the basic query starts as getting all "observers" who share three "subjects" with a given observer:
select f.observerid
from followers f join
followers f2
on f.subjectid = f2.subjectid and
f2.observerid = 2
group by f.observerid
having count(*) = 3;
The rest of the query is just joining in the names to fit into your paradigm of using names for references rather than ids.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Consider the following refactored SQL code (untested without data) for use in stored procedure.
select u.`name`
from `User` u
inner join
(select subf.observerID, subf.subjectID
from follower subf
where subf.observerID <> xuid
) f
on u.UID = f.subjectID
inner join
(select f1.observerID
from follower f1
inner join follower f2
on f1.subjectID = f2.subjectID
and f1.observerID <> xuid
and f2.observerID = xuid
group by f1.observerID
having count(*) >= 3
) o
on f.observerID = o.observerID
add a comment |
up vote
0
down vote
Consider the following refactored SQL code (untested without data) for use in stored procedure.
select u.`name`
from `User` u
inner join
(select subf.observerID, subf.subjectID
from follower subf
where subf.observerID <> xuid
) f
on u.UID = f.subjectID
inner join
(select f1.observerID
from follower f1
inner join follower f2
on f1.subjectID = f2.subjectID
and f1.observerID <> xuid
and f2.observerID = xuid
group by f1.observerID
having count(*) >= 3
) o
on f.observerID = o.observerID
add a comment |
up vote
0
down vote
up vote
0
down vote
Consider the following refactored SQL code (untested without data) for use in stored procedure.
select u.`name`
from `User` u
inner join
(select subf.observerID, subf.subjectID
from follower subf
where subf.observerID <> xuid
) f
on u.UID = f.subjectID
inner join
(select f1.observerID
from follower f1
inner join follower f2
on f1.subjectID = f2.subjectID
and f1.observerID <> xuid
and f2.observerID = xuid
group by f1.observerID
having count(*) >= 3
) o
on f.observerID = o.observerID
Consider the following refactored SQL code (untested without data) for use in stored procedure.
select u.`name`
from `User` u
inner join
(select subf.observerID, subf.subjectID
from follower subf
where subf.observerID <> xuid
) f
on u.UID = f.subjectID
inner join
(select f1.observerID
from follower f1
inner join follower f2
on f1.subjectID = f2.subjectID
and f1.observerID <> xuid
and f2.observerID = xuid
group by f1.observerID
having count(*) >= 3
) o
on f.observerID = o.observerID
answered Nov 11 at 0:33
Parfait
48k84066
48k84066
add a comment |
add a comment |
up vote
0
down vote
I think the basic query starts as getting all "observers" who share three "subjects" with a given observer:
select f.observerid
from followers f join
followers f2
on f.subjectid = f2.subjectid and
f2.observerid = 2
group by f.observerid
having count(*) = 3;
The rest of the query is just joining in the names to fit into your paradigm of using names for references rather than ids.
add a comment |
up vote
0
down vote
I think the basic query starts as getting all "observers" who share three "subjects" with a given observer:
select f.observerid
from followers f join
followers f2
on f.subjectid = f2.subjectid and
f2.observerid = 2
group by f.observerid
having count(*) = 3;
The rest of the query is just joining in the names to fit into your paradigm of using names for references rather than ids.
add a comment |
up vote
0
down vote
up vote
0
down vote
I think the basic query starts as getting all "observers" who share three "subjects" with a given observer:
select f.observerid
from followers f join
followers f2
on f.subjectid = f2.subjectid and
f2.observerid = 2
group by f.observerid
having count(*) = 3;
The rest of the query is just joining in the names to fit into your paradigm of using names for references rather than ids.
I think the basic query starts as getting all "observers" who share three "subjects" with a given observer:
select f.observerid
from followers f join
followers f2
on f.subjectid = f2.subjectid and
f2.observerid = 2
group by f.observerid
having count(*) = 3;
The rest of the query is just joining in the names to fit into your paradigm of using names for references rather than ids.
answered Nov 11 at 12:32
Gordon Linoff
744k32285390
744k32285390
add a comment |
add a comment |
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Why do you input
in_name
as Char. You must be having its integer ID value also. My suggestion is to use that instead. It will be more performant. Moreoever, what happens if there are two users with same name(s) ?– Madhur Bhaiya
Nov 11 at 8:22
Also, do you have access to latest version of MySQL (version 8.0.2 and above) ?
– Madhur Bhaiya
Nov 11 at 8:26