Mysql – Finding Records that Have Same Options

MySQLrelational-division

Trying to figure out how find out which user records have at least 1 similar option enabled and then display these.

Tables:

user (id, username)
thing (id,title)
uthing (id, uid, tid)

Uthing table has a record of what options each user have selected.

uid = user.id
tid = thing.id

I need to compare the options (uthing) from user id 1 with all other users.

How can find users that have 1 or more options(uthing) of USER ID 1 and return these users with the options found?

This query will give me all options for user id 1.

select thing.title,thing.id as thingid 
from thing join uthing on uthing.tid = thing.id where uthing.uid = 1;

Best Answer

You achieve this by following this query

select uthing.id, uthing.uid, thing.title
from uthing left join thing on uthing.tid = thing.id
where uthing.tid in (
        select thing.id
        from thing join uthing on uthing.tid = thing.id
        where uthing.uid = 1
     );

Related Solutions

Mysql – Help with Query. Finding records that have the same relationships (MySQL)

SELECT B.name
FROM
(
    SELECT BB.listing_id id,COUNT(1) taxon_count
    FROM
    (
        SELECT id taxon_id FROM taxons
        WHERE name IN ('Ford','Exhaust')
    ) AA
    INNER JOIN listings_taxons BB
    USING (taxon_id)
    GROUP BY listing_id HAVING COUNT(1) = 2
) A
INNER JOIN listings B USING (id);

Subquery A will bring back all listing_ids that have Ford, Exhaust, or both. Doing the GROUP BY count within Subquery A gives any listing id that has a COUNT(1) of 2 has both Ford and Exhaust taxon ids becasue BB.listing_id would appears twice thus HAVING COUNT(1) = 2. Then Subquery A has an INNER JOIN with listings.

Make sure you have the following indexes

ALTER TABLE listings_taxons ADD INDEX taxon_listing_ndx (taxon_id,listing_id);
ALTER TABLE taxons ADD INDEX name_id_ndx (name,id);

Here is some sample data

drop database if exists nwwatson;
create database nwwatson;
use nwwatson
create table listings
(id int not null auto_increment,
name varchar(25),
primary key (id),
key (name));
create table taxons like listings;
create table listings_taxons
(
    listing_id int,
    taxon_id int,
    primary key (listing_id,taxon_id),
    unique key (taxon_id,listing_id)
);
insert into listings (name) values ('SteeringWheel'),('WindShield'),('Muffler'),('AC');
insert into taxons (name) values ('Ford'),('Escort'),('Buick'),('Exhaust'),('Mustard');
insert into listings_taxons values
(1,1),(1,3),(1,5),(2,1),(2,2),(2,3),(2,5),
(3,1),(3,4),(4,2),(4,3),(4,4),(5,1),(5,5);
SELECT * FROM listings;
SELECT * FROM taxons;
SELECT * FROM listings_taxons;
SELECT B.name
FROM
(
    SELECT BB.listing_id id,COUNT(1) taxon_count
    FROM
    (
        SELECT id taxon_id FROM taxons
        WHERE name IN ('Ford','Exhaust')
    ) AA
    INNER JOIN listings_taxons BB
    USING (taxon_id)
    GROUP BY listing_id HAVING COUNT(1) = 2
) A
INNER JOIN listings B USING (id);

Here is it executed

mysql> drop database if exists nwwatson;
Query OK, 3 rows affected (0.09 sec)

mysql> create database nwwatson;
Query OK, 1 row affected (0.00 sec)

mysql> use nwwatson
Database changed
mysql> create table listings
    -> (
    -> id int not null auto_increment,
    -> name varchar(25),
    -> primary key (id),
    -> key (name)
    -> );
Query OK, 0 rows affected (0.08 sec)

mysql> create table taxons like listings;
Query OK, 0 rows affected (0.05 sec)

mysql> create table listings_taxons
    -> (
    ->     listing_id int,
    ->     taxon_id int,
    ->     primary key (listing_id,taxon_id),
    ->     unique key (taxon_id,listing_id)
    -> );
Query OK, 0 rows affected (0.08 sec)

mysql> insert into listings (name) values ('SteeringWheel'),('WindShield'),('Muffler'),('AC');
Query OK, 4 rows affected (0.06 sec)
Records: 4  Duplicates: 0  Warnings: 0

mysql> insert into taxons (name) values ('Ford'),('Escort'),('Buick'),('Exhaust'),('Mustard');
Query OK, 5 rows affected (0.06 sec)
Records: 5  Duplicates: 0  Warnings: 0

mysql> insert into listings_taxons values
    -> (1,1),(1,3),(1,5),(2,1),(2,2),(2,3),(2,5),
    -> (3,1),(3,4),(4,2),(4,3),(4,4),(5,1),(5,5);
Query OK, 14 rows affected (0.11 sec)
Records: 14  Duplicates: 0  Warnings: 0

mysql> SELECT * FROM listings;
+----+---------------+
| id | name          |
+----+---------------+
|  4 | AC            |
|  3 | Muffler       |
|  1 | SteeringWheel |
|  2 | WindShield    |
+----+---------------+
4 rows in set (0.00 sec)

mysql> SELECT * FROM taxons;
+----+---------+
| id | name    |
+----+---------+
|  3 | Buick   |
|  2 | Escort  |
|  4 | Exhaust |
|  1 | Ford    |
|  5 | Mustard |
+----+---------+
5 rows in set (0.00 sec)

mysql> SELECT * FROM listings_taxons;
+------------+----------+
| listing_id | taxon_id |
+------------+----------+
|          1 |        1 |
|          1 |        3 |
|          1 |        5 |
|          2 |        1 |
|          2 |        2 |
|          2 |        3 |
|          2 |        5 |
|          3 |        1 |
|          3 |        4 |
|          4 |        2 |
|          4 |        3 |
|          4 |        4 |
|          5 |        1 |
|          5 |        5 |
+------------+----------+
14 rows in set (0.00 sec)

mysql> SELECT B.name
    -> FROM
    -> (
    ->     SELECT BB.listing_id id,COUNT(1) taxon_count
    ->     FROM
    ->     (
    ->         SELECT id taxon_id FROM taxons
    ->         WHERE name IN ('Ford','Exhaust')
    ->     ) AA
    ->     INNER JOIN listings_taxons BB
    ->     USING (taxon_id)
    ->     GROUP BY listing_id HAVING COUNT(1) = 2
    -> ) A
    -> INNER JOIN listings B USING (id);
+---------+
| name    |
+---------+
| Muffler |
+---------+
1 row in set (0.00 sec)

mysql>

Give it a Try !!!

Mysql – Deleting the MySQL ‘root’ user on purpose

You can rename the 'root' user.

To rename the administrator’s username, use the rename command in the MySQL console:

mysql> RENAME USER root TO new_user;

The MySQL “RENAME USER” command first appeared in MySQL version 5.0.2. If you use an older version of MySQL, you can use other commands to rename a user:

mysql> use mysql;

mysql> update user set user=”new_user” where user=”root”;

mysql> flush privileges;

From here.

Best Answer

Related Solutions

Mysql – Help with Query. Finding records that have the same relationships (MySQL)

Mysql – Deleting the MySQL ‘root’ user on purpose

Related Question