Is it possible to link product id's from different tables to one universal product id? eg 1014 id from table A and 2015 id from table B to one universal 10 id in table C ?
In this case you could do something like this:
First your internal products:
master_id, name, description, etc...
1, "Keyboard", "Nice"
2, "Mouse", "Microsoft"
3, "Monitor", "Bright"
4, "Printer", "Not the best"
Second table a and table b would have a master_id column that references one of those ids.
Then to select all keyboards from table a or table b:
SELECT * FROM table_a JOIN products ON table_a.master_id =
products.master_id WHERE products.master_id =1;
SELECT * FROM table_b JOIN products ON table_a.master_id =
products.master_id WHERE products.master_id =1;
you can then get all keyboards from BOTH tables via union:
SELECT * FROM table_a JOIN products ON table_a.master_id =
products.master_id WHERE products.master_id =1 UNION
SELECT * FROM table_b JOIN products ON table_a.master_id =
products.master_id WHERE products.master_id =1;
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Related
I'm currently using StandardSQL in BigQuery, I tried to join two sets of table one of which is a pseudo-column table partitioned by day.
I tried to use this query below:
SELECT
DISTINCT DATE(create_time) AS date,
user_id,
city_name,
transaction_id,
price
FROM
table_1 a
LEFT JOIN (SELECT user_id, city_name FROM table_2) b
ON (a.user_id = b.user_id AND DATE(create_time) = _PARTITIONDATE)
I've tried this kind of JOIN (using _PARTITIONDATE) and worked out, but for this particular query I got an error message:
Unrecognized name: _PARTITIONDATE
Can anyone tell me why this happened, and how could I solve this? Thanks in advance.
The issue is that you are not selecting the _PARTITIONDATE field from table_2 when joining it so it can't recognize it:
SELECT user_id, city_name FROM table_2
In order to solve it you can add it as follows:
SELECT
DISTINCT DATE(create_time) AS date,
user_id,
city_name,
transaction_id,
price
FROM
table_1 a
LEFT JOIN (SELECT _PARTITIONDATE AS pd, user_id, city_name FROM table_2) b
ON (a.user_id = b.user_id AND DATE(create_time) = pd)
Note that you'll need an alias such as pd as it's a pseudocolumn
Probably it was working in the past if you were joining two tables directly such as in (you don't get selectivity benefits in that case):
FROM
table_1 a
LEFT JOIN table_2 b
ON (a.user_id = b.user_id AND DATE(create_time) = _PARTITIONDATE)
I would like to join tables . Could you please help?
Select Number, OwnerId from DNIS.numbers
select ID,Name from DNIS.owners
Thank you.
Normally, SQL servers allow you to join tables from different databases as long as the former all belong to them. Here is an example showing you how to do this (all you have to do is to explicitly write the database names associated to each table in the query):
SELECT N.Number, N.OwnerId, O.ID, O.Name
FROM DB1.[dbo].DNIS numbers N
JOIN DB2.[dbo].DNIS owners O ON O.ID = N.OwnerId
You can also use the following syntax:
SELECT N.Number, N.OwnerId, O.ID, O.Name
FROM DB1..DNIS numbers N
JOIN DB2..DNIS owners O ON O.ID = N.OwnerId
In order to accomplish that you will have to specify the table and column names in your join statement, like so:
SELECT db1.tablename.column, db2.tablename.column
FROM db1.tablename INNER JOIN db2.tablename
ON db1.tablename.id = db2.tablename.id;
Issue:
You are given three tables: Students, Friends and Packages.
Students contains two columns: ID and Name.
Friends contains two columns: ID and Friend_ID (ID of the ONLY best friend).
Packages contains two columns: ID and Salary (offered salary in $ thousands per month).
Write a query to output the names of those students whose best friends got offered a higher salary than them. Names must be ordered by the salary amount offered to the best friends. It is guaranteed that no two students got same salary offer.
Code:
This is the code that I have come up with but it does not produce correct results. Can anyone let me know why?
select TableA.name
from
(select s.id,s.name,p.salary from students s inner join packages p on s.id=p.id) TableA,
(select f.id,f.friend_id, p2.salary from friends f inner join packages p2 on f.friend_id=p2.id) TableB
where TableA.id=TableB.id And TableA.salary>TableB.salary
order by TableB.salary desc;
I think in your query you wrote AND TableA.salary < TableB.salary instead of AND TableA.salary > TableB.salary.
Moreover I think your query can be written in a more synthetic way.
On MSSQL (but it works on MYSQL too, as query is very basic), you can try to use this one:
SELECT s.id
,s.NAME
,p.salary
, f.friend_id, p2.salary as friend_salary
FROM students s
INNER JOIN packages p ON s.id = p.id
LEFT JOIN friends f ON f.id = s.id
LEFT JOIN packages p2 ON f.friend_id = p2.id
WHERE p.salary <= p2.salary
ORDER BY s.id;
Output:
id NAME salary friend_id friend_salary
1 John 1000 2 1200
3 Pete 800 1 1000
Sample data:
CREATE TABLE students (id int, NAME VARCHAR(30));
CREATE TABLE packages (id int, salary INT);
CREATE TABLE friends (id int, friend_id INT);
INSERT INTO students values (1,'John');
INSERT INTO students values (2,'Arthur');
INSERT INTO students values (3,'Pete');
INSERT INTO packages values (1,1000);
INSERT INTO packages values (2,1200);
INSERT INTO packages values (3,800);
INSERT INTO friends values (1,2);
INSERT INTO friends values (2,3);
INSERT INTO friends values (3,1);
I used CTE for easy code readability. I am not sure whether it is fully optimized or not. But, it yields the result as expected from the question.
with std_salary as (
SELECT s.id, s.name, p.salary
FROM Students s
JOIN Packages p
ON s.id=p.id),
friend_salary as (
SELECT f.id, p.salary
FROM Friends f
JOIN Packages p
ON f.friend_id=p.id
)
SELECT name
FROM
(SELECT std_salary.name, std_salary.salary as own, friend_salary.salary as friend
FROM std_salary
JOIN friend_salary
ON std_salary.id=friend_salary.id) as final
WHERE final.own<final.friend
ORDER BY final.friend;
This worked for me in MS SQL
SELECT a.name
FROM (SELECT students.id as main_id, students.name, packages.salary
FROM students join packages on students.id = packages.id) a
JOIN (SELECT f.id as main_id1, p.salary
FROM friends f JOIN packages p ON f.friend_id = p.id) b
ON a.main_id = b.main_id1
WHERE b.salary>a.salary
ORDER BY b.salary ASC;
you have written 'where TableA.salary>TableB.salary' implying that you want to find rows where your salary is > than your friends. But the question asked was the opposite (to find names where the firends salary is > than your salary) so you can change that to 'where TableB.salary>TableA.salary' and it should work.
select my_name from
(select s.id as my_id,s.name my_name,p.salary as my_salary from students s
inner join packages p on s.id=p.id) as my_tbl inner join (select f.id as
id,f.friend_id as frnd_id,p.salary as frnd_salary from friends f inner join
packages p on f.friend_id=p.id ) as frnd_tbl on my_id=id where
frnd_salary>my_salary order by frnd_salary;
I have five tables: users, interests, animals, interests_animals, interests_users.
User
foo
Interest 1, 2, 3
Animal 1, 2, 3
foo has interests 1, 2, 3
Interest 1 has Animal 1, 2
Interest 2 has Animal 1, 2, 3
Interest 3 has Animal 3
I need return all animals through interests grouped for interest id of foo ordered by animals count
I trying like this:
SELECT animals.* FROM animals
INNER JOIN interests_animals ON animals.id = interests_animals.animal_id
INNER JOIN interests ON interests_animals.interest_id = interests.id
INNER JOIN interests_users ON interests.id = interests_users.interest_id
WHERE interests_users.user_id = XXX
GROUP BY animals.id
ORDER BY COUNT(interests_animals.animal_id);
I need that the animals are returned in orders 2, 1, 3, but always returning 1,2,3
You need explicitly specify column(s), on which you do GROUP BY in SELECT clause.
All other parts of SELECT clause must be aggregates like count(), sum(), etc.
Notice, that we use count(distinct ..) here because each animal ID might appear multiple times due to the chain of JOINs:
SELECT
interests.id,
COUNT(DISTINCT animals.id) as animals_count
JOIN interests_animals ON animals.id = interests_animals.animal_id
JOIN interests ON interests_animals.interest_id = interests.id
JOIN interests_users ON interests.id = interests_users.interest_id
WHERE interests_users.user_id = XXX
GROUP BY 1
ORDER BY 2 desc;
-- in GROUP BY and ORDER BY, it is usually convenient to use just numbers -- "1" means "the 1st column of SELECT clause", etc.
Also, "INNER" is an optional keyword (simply "JOIN" and "INNER JOIN" are the same thing).
Also, as a side note, you might found useful to add this to your SELECT clause:
, array_agg(animals.id order by animals.id) as animal_ids
-- this will give you integer array of all animal IDs that relate to a particular interest, ordered.
Database is Teradata
I have two table which I am trying to join. Following are the table structures. When I join these table I expect to get two rows as output but getting 4 rows.what is reason for this behavior. Join based on three keys should uniquely identify a row but still getting 4 rows as output. Any help is appreciated.
TableA
Weekkey|segment|type|users
201501|1|A|100
201501|1|B|100
TableB
Weekkey|segment|type|revenue
201501|1|A|200
201501|1|B|200
when I join these two table using the following query i get the following result
select a.* ,b.user
from tablea a left join tableb b on a.weekkey=b.weekkey
and a.segment=b.segment
and a.type=b.type
Weekkey|segment|type|revenue|users
201501|1|A|200|100
201501|1|B|200|100
201501|1|A|200|100
201501|1|B|200|100
Using sql server, here is ddl and sample data along with the query you posted. The output you state you are getting doesn't happen here.
create table #tablea
(
Weekkey int
, segment int
, type char(1)
, users int
)
insert #tablea
select 201501, 1, 'A', 100 union all
select 201501, 1, 'B', 100
create table #TableB
(
Weekkey int
, segment int
, type char(1)
, revenue int
)
insert #TableB
select 201501, 1, 'A', 200 union all
select 201501, 1, 'B', 200
select a.*
, b.revenue
from #tablea a
left join #tableb b on a.weekkey = b.weekkey
and a.segment = b.segment
and a.type = b.type
drop table #tablea
drop table #TableB