many-to-many and many-to-many intersections - mysql

Say I have a database that has people, grocery stores, and items you can buy in the store, like so:
Stores People Foods
----------------- ------------------ ------------------
| id | name | | id | name | | id | name |
----------------- ------------------ ------------------
| 1 | Giant | | 1 | Jon Skeet | | 1 | Tomatoes |
| 2 | Vons | | 2 | KLee1 | | 2 | Apples |
| 3 | Safeway | ------------------ | 3 | Potatoes |
----------------- ------------------
I have an additional table which keep track of which stores sell what:
Inventory
--------------------
| store_id| food_id|
--------------------
| 1 | 1 |
| 1 | 2 |
| 2 | 1 |
| 3 | 1 |
| 3 | 2 |
| 3 | 3 |
--------------------
And I have another table that has shopping lists on it
Lists
---------------------
| person_id| food_id|
---------------------
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 3 |
---------------------
My question is, given a person, or their id, what is the best way to figure out what stores they can go to so they will get everything on their list. Is there a pattern for these types of computations in MySQL?
My attempt (very ugly and messy) is something like:
-- Given that _pid is the person_id we want to get the list of stores for.
SELECT stores.name, store_id, num, COUNT(*) AS counter
FROM lists
INNER JOIN inventory
ON (lists.food_id=inventory.food_id)
INNER JOIN (SELECT COUNT(*) AS num
FROM lists WHERE person_id=_pid
GROUP BY person_id) AS T
INNER JOIN stores ON (stores.id=store_id)
WHERE person_id=_pid
GROUP BY store_id
HAVING counter >= num;
Thanks for your time!
Edit SQL Fiddle with Data

If I were to solved the problem, I'll join the four tables with their linking column (specifically the foreign keys) then a subquery on the HAVING clause to count the number of items on the list for each person. Give this a try,
SET #personID := 1;
SELECT c.name
FROM Inventory a
INNER JOIN Foods b
ON a.food_id = b.id
INNER JOIN Stores c
ON a.store_id = c.id
INNER JOIN Lists d
ON d.food_id = b.id
WHERE d.person_id = #personID
GROUP BY c.name
HAVING COUNT(DISTINCT d.food_id) =
(
SELECT COUNT(*)
FROM Lists
WHERE person_ID = #personID
)
SQLFiddle Demo

#JohnWoo: why DISTINCT?
Another one...
SET #pid=2;
SELECT store_id, name
FROM inventory
JOIN lists ON inventory.food_id=lists.food_id
JOIN stores ON store_id=stores.id
WHERE person_id=#pid
GROUP BY store_id
HAVING COUNT(*)=(
SELECT COUNT(*)
FROM lists
WHERE person_id=#pid
);

Related

SQL/MySQL - Select and return array column on one-to-many table join [duplicate]

We have 3 tables :
donations
purposes
expenses
Donations :
+--------+------+
| do_id | name |
+--------+------+
| 1 | A |
| 2 | B |
| 3 | A |
| 4 | D |
| 5 | B |
| 6 | B |
| 7 | A |
| 8 | B |
+--------+----- +
purposes:
+-------+-------+--------+
| pu_id | do_id | purpose|
+-------+-------+--------+
| 1 | 2 | abc |
| 2 | 2 | def |
| 3 | 2 | gih |
| 4 | 3 | jkl |
+-------+-------+--------+
expense :
+-------+-------+---------+
| ex_id | do_id | expense |
+-------+-------+---------+
| 1 | 2 | abc |
| 2 | 2 | def |
| 3 | 2 | gih |
| 4 | 3 | jkl |
+-------+-------+---------+
Now i want to make query to get all donations for donor B and join purposes table to get all purposes related to every donation_id then join expenses table to get all expenses related to donation_id and put all of that in every loop independently something like that
Row number 0
donation_id = 1
array(purposes)
array(expenses)
Row number 1
donation_id = 2
array(purposes)
array(expenses)
Row number 2
donation_id = 3
array(purposes)
array(expenses)
Row number 3
donation_id = 4
array(purposes)
array(expenses)
This is my try :
SELECT *, (
SELECT *
FROM `donation_purposes`
WHERE `donation_purposes`.`dopu_donation_id` = 4
) AS `purposes`
FROM `donations`
WHERE `donation_id` = '4'
thanks in advance
You should be able to solive this with an aggregate query using MySQL aggregate function JSON_ARRAYAGG(), like :
SELECT
d.do_id,
JSON_ARRAYAGG(p.purpose) purposes,
JSON_ARRAYAGG(e.expense) expenses
FROM donations d
INNER JOIN purposes p ON p.do_id = d.do_id
INNER JOIN expense e ON e.do_id = d.do_id
GROUP BY d.do_id
I you want to avoid duplicate values in the array, and as JSON_ARRAYAGG() (sadly) does not support the DISTINCT option, you can move aggregation to subqueries, like :
SELECT
d.do_id,
p.agg purpose,
e.agg expenses
FROM donations d
INNER JOIN (
SELECT do_id, JSON_ARRAYAGG(purpose) agg FROM purposes GROUP BY do_id
) p ON p.do_id = d.do_id
INNER JOIN (
SELECT do_id, JSON_ARRAYAGG(expense) agg FROM expense GROUP BY do_id
) e ON e.do_id = d.do_id
This demo on DB Fiddle returns :
| do_id | purpose | expenses |
| ----- | --------------------- | --------------------- |
| 2 | ["abc", "def", "gih"] | ["abc", "def", "gih"] |
| 3 | ["jkl"] | ["jkl"] |
1st Select Query Purposes
SELECT purposes.* FROM purposes
LEFT JOIN donations
ON purposes.do_id = donations.do_id
WHERE donations.do_id = '2' //This depends on the id of the donation
ORDER BY purposes.do_id ASC
2nd Select Query Expenses
SELECT expense.* FROM expense
LEFT JOIN donations
ON expense.do_id = donations.do_id
WHERE donations.do_id = '2' //This depends on the id of the donation
ORDER BY expense.ex_id ASC
All queries generated are from the table structure you've provided, but your question is quite vague!!

Selecting values from second column alongside the values from first column in the same row

I am trying to get values matching the value from the second column. For example, I want to know who is the sender for Bill Gates by only using IDs.
I have two tables,
*users* table
| user_ID | Full_name |
| -------- | -------------- |
| 1 | Steve Jobs |
| 2 | Bill Gates |
| 3 | Elon Musk |
*relationships* table (with both column foreign keys)
| user_sender | user_receiver |
| ------------ | -------------- |
| 1 | 2 |
| 3 | 1 |
| 3 | 2 |
I want to select based on "user_receiver" column the matching values in the column "user_sender"
For example, I want to know who is user_sender for 2
OUTPUT:
| | |
| ------------ | -------------- |
| 1 | 2 |
| 3 | 2 |
You need to join the tables and select the rows you want
you have access to all columns of both tables by addressing them with their alias
SELECT u.user_ID , u.Full_name,r.user_receiver
FROM users u JOIN
relationships r ON u.user_ID = r.user_sender
WHERE r.user_receiver = 2
If you want to look based on the name, then join the relationships to users.
SELECT
rel.user_sender
, rel.user_receiver
-- , sender.Full_name AS sender_name
-- , receiver.Full_name AS receiver_name
FROM relationships AS rel
JOIN users AS sender ON sender.user_ID = rel.user_sender
JOIN users As receiver ON receiver.user_ID = rel.user_receiver
WHERE receiver.Full_name = 'Bill Gates'
If you already know the user_receiver number, and you only want the ID's
SELECT *
FROM relationships
WHERE user_receiver = 2

how to perform an outer join in mysql

I have a table A that contains tree columns, id, users ids and vehicle id. And a table B that contains vehicleid, and vehicle name.
Table A
---------------------------
| Id | User_id |Vehicle_id|
---------------------------
| 1 | 1 | 2 |
| 2 | 1 | 3 |
| 3 | 1 | 4 |
| 4 | 2 | 2 |
| 5 | 2 | 3 |
| 6 | 4 | 5 |
---------------------------
Table B
-------------------
| Id |Vehicle_name|
-------------------
| 1 | Car |
| 2 | Bike |
| 3 | Plane |
| 4 | Boat |
| 5 | Rocket |
-------------------
Given a user id, I need to get all vehicle names, that doesn't match with table A. I've tried Outer joins, but I can't manage to do get the info that i need.
For example: Given user id 1, the query should return Car and Rocket.
thanks in advance
This is simple enough using not in or not exists:
select b.*
from b
where not exists (select 1
from a
where a.vehicle_id = b.id and a.user_id = #a_user_id
);
I also thought of using a cross join and was able to get the output in case you are more comfortable with join logic.
SELECT CJOIN.USER_ID, CJOIN.VEHICLE_ID, CJOIN.VEHICLE_NAME
FROM
(SELECT DISTINCT A.USER_ID, B.ID AS VEHICLE_ID, B.VEHICLE_NAME FROM TABLE_A A CROSS JOIN TABLE_B B) CJOIN
LEFT JOIN
TABLE_A D
ON CJOIN.USER_ID = D.USER_ID AND CJOIN.VEHICLE_ID = D.VEHICLE_ID
WHERE D.USER_ID IS NULL AND D.VEHICLE_ID IS NULL;
First, I got all possible combinations of USER_ID x VEHICLE_ID by a cross join and used this table in a left join to pull records for which there is no match.

MySQL - join multiple mapped tables and count records with different mapping conditions

It's the 3rd day I'm trying to write a MySQL query. Did lots of search, but it still doesn't work as expected. I'll try to simplify tables as much as possible
System has tkr_restaurants table:
restaurant_id | restaurant_name
1 | AA
2 | BB
3 | CC
Each restaurant has a division assigned (tkr_divisions table):
division_id | restaurant_id | division_name
1 | 1 | AA-1
2 | 1 | AA-2
3 | 2 | BB-1
Then there are meals in tkr_meals_to_restaurants_divisions table, where each meal can be assigned (mapped) to whole restaurant(s) and/or specific division(s). If meal is mapped to restaurant, all restaurant's divisions should see it. If meal is mapped to division(s), only specific division(s) should see it.
meal_id | mapped_restaurant_id | mapped_division_id
1 | 1 | NULL
2 | NULL | 1
3 | NULL | 2
I need to display a list of restaurants and number of meals mapped to it depending on user permissions.
Example 1: if user has permissions to access whole restaurant_id 1 and restaurant_3 (and no specific divisions), then list should be:
AA | 3
CC | 0
(because user can access meals mapped to restaurant 1 + all its division, and restaurant 3 + all its divisions (even if restaurant 3 has no divisions/meals mapped))
Example 2: if user has permissions to access only division_id 1, then list should be:
AA | 1
(because user can only access meals mapped to division 1).
The closest query I could get is:
Example 1:
SELECT *,
(SELECT COUNT(DISTINCT meal_id)
FROM
tkr_meals_to_restaurants_divisions
WHERE
tkr_meals_to_restaurants_divisions.mapped_restaurant_id=tkr_restaurants.restaurant_id
OR tkr_meals_to_restaurants_divisions.mapped_division_id=tkr_divisions.division_id)AS total_meals
FROM
tkr_restaurants
LEFT JOIN
tkr_divisions
ON tkr_restaurants.restaurant_id=tkr_divisions.restaurant_id
WHERE
tkr_restaurants.restaurant_id IN (1, 3)
OR tkr_restaurants.restaurant_id IN (
SELECT restaurant_id
FROM tkr_divisions
WHERE division_id IN (NULL)
)
GROUP BY
tkr_restaurants.restaurant_id
ORDER BY
tkr_restaurants.restaurant_name
However, result was:
AA | 2
CC | 0
I believe I'm greatly over-complicating this query, but all the simpler queries I wrote produced even more inaccurate results.
What about this query:
SELECT
FROM tkr_restaurants AS a
JOIN tkr_divisions AS b
ON a.restaurant_id = b.restaurant_id
LEFT OUTER JOIN tkr_meals_to_restaurants_divisions AS c
ON (c.mapped_restaurant_id = a.restaurant_id OR c.mapped_division_id = b.division_id)
As a Base four your further work. It combine all information into one table. If you add e.g. this:
WHERE a.restaurant_id IN (1, 3)
the result will be
| restaurant_id | restaurant_name | division_id | restaurant_id | division_name | meal_id | mapped_restaurant_id | mapped_division_id |
|---------------|-----------------|-------------|---------------|---------------|---------|----------------------|--------------------|
| 1 | AA | 1 | 1 | AA-1 | 1 | 1 | (null) |
| 1 | AA | 2 | 1 | AA-2 | 1 | 1 | (null) |
| 1 | AA | 1 | 1 | AA-1 | 2 | (null) | 1 |
| 1 | AA | 2 | 1 | AA-2 | 3 | (null) | 2 |
just count the distinct meal ids with COUNT(DISTINCT c.meal_id) and take the restaurant name to get AA: 3 for your example 2
I used a sqlfiddle: http://sqlfiddle.com/#!9/fa2b78/18/0
[EDIT]
Change JOIN tkr_divisions AS b to LEFT OUTER JOIN tkr_divisions AS b
Change SELECT * to SELECT a.restaurant_name, COUNT(DISTINCT c.meal_id)
Add a GROUP BY a.restaurant_name at the end.
Update the SQL Fiddle (new link)

Join on same column name

Hello there I want to get data from two tables that share same column name. My table structure are
Table patients
---------------------------------------
| id | affiliate_id | somecolumn |
---------------------------------------
| 1 | 8 | abc |
---------------------------------------
| 2 | 8 | abc |
---------------------------------------
| 3 | 9 | abc |
---------------------------------------
Table Leads
---------------------------------------
| id | affiliate_id | someothern |
---------------------------------------
| 1 | 8 | xyz |
---------------------------------------
| 2 | 8 | xyz |
---------------------------------------
| 3 | 3 | xyz |
---------------------------------------
Now my requirement was to get COUNT(ID) from both tables in a single query. I want result like
----------------------------------------------------
| affiliate_id | total_patients | total_leads |
----------------------------------------------------
| 8 | 2 | 2 |
----------------------------------------------------
| 9 | 1 | 0 |
----------------------------------------------------
| 3 | 0 | 1 |
----------------------------------------------------
I wrote following query
SELECT `p`.`affiliate_id`, COUNT(p.id) AS `total_patients`,
COUNT(cpl.id) AS `total_leads`
FROM `patients` AS `p`
INNER JOIN `leads` AS `cpl` ON p.affiliate_id =cpl.affiliate_id
GROUP BY `p`.`affiliate_id`
But I am not getting result . This query results giving only one affiliate with same number of total_patients and total_leads
The problem is that you need to get a list of the distinct affiliate_id first and then join to your other tables to get the result:
select a.affiliate_id,
count(distinct p.id) total_patients,
count(distinct l.id) total_leads
from
(
select affiliate_id
from patients
union
select affiliate_id
from leads
) a
left join patients p
on a.affiliate_id = p.affiliate_id
left join leads l
on a.affiliate_id = l.affiliate_id
group by a.affiliate_id;
See SQL Fiddle with Demo
Two ways:
Select l.affiliate_id ,
count(distinct p.id) patientCount,
count(distinct l.id) LeadCOunt
From patients p Join leads l
On l.affiliate_id = p.Affiliate_id
Group By l.affiliate_id
or, (assuming affiliates are in their own table somewhere)
Select Affiliate_id,
(Select Count(*) From Patients
Where Affiliate_id = a.Affiliate_id) patientCount,
(Select Count(*) From Leads
Where Affiliate_id = a.Affiliate_id) LeadCount
From affiliates a