SQL Union SELECT from two tables then combine to one row - mysql

I have two tables (tests and reserved) which reference each other it is a 1:1 relationship.
One test's ID is reserved for another test's ID which is in the same table but kept in the reserved table as a look up.
Here's some sample data:
tests:
test_id | summary_id | ref
1 | 1 | ref1
2 | 2 | ref2
reserved:
reserved_id | reserved_summary_id | reserved_for_summary_id
1 | 1 | 2
Currently I am using a UNION to get them both:
SELECT * FROM reserved r, tests t WHERE t.summary_id = r.reserved_summary_id
UNION
SELECT * FROM reserved r, tests t WHERE t.summary_id = r.reserved_for_summary_id
I have a fiddle here.
How can I combine that query into one row? I have tried this one from SO. But get an error.

Are you trying to JOIN the summary table twice with the the tests table so you can access the details of each reserved_summary and reserved_for_summary ?
If yes, then you need :
SELECT
r.reserved_id,
t1.ref ref_of_reserved_summary_id,
t2.ref ref_of_reserved_for_summary_id
FROM
reserved r
INNER JOIN tests t1 ON t1.summary_id = r.reserved_summary_id
INNER JOIN tests t2 ON t2.summary_id = r.reserved_for_summary_id
This will return something like :
reserved_id | ref_of_reserved_summary_id | ref_of_reserved_for_summary_id
1 | ref1 | ref2
You can switch to LEFT JOIN to avoid filtering out records where one of the summaries is not registered in the tests table.

Related

How to find all the opposite combinations between two columns in SQL

I am making a web dating app that needs to match users and let them chat with each other.
I want to figure out how to find all the matches for a particular user.
Right now I have a table called follows that has 2 columns.
UserID | MatchUserID
--------------------
1 | 2
2 | 1
1 | 3
1 | 4
1 | 5
4 | 1
5 | 4
The idea is that for two users to match they need to follow one another. The table above shows which user follows which.
Assuming that the user who is currently logged on is UserID = 1.
I need a query that will return from the MatchUserID table the following results:
2, 4
In a way, I am looking to find all the opposite combinations between the two columns.
This is the code I use to create the table.
CREATE TABLE Match
(
UserID INT NOT NULL,
MatchUserID INT NOT NULL,
PRIMARY KEY (UserID, MatchUserID)
);
You can do it with a self join:
select m.MatchUserID
from `Match` m inner join `Match` mm
on mm.MatchUserID = m.UserId
where
m.UserId = 1
and
m.MatchUserID = mm.UserId
See the demo.
Results:
| MatchUserID |
| ----------- |
| 2 |
| 4 |
The simplest way possibly is to use EXISTS and a correlated subquery that searches for the other match.
SELECT t1.matchuserid
FROM elbat t1
WHERE t1.userid = 1
AND EXISTS (SELECT *
FROM elbat t2
WHERE t2.matchuserid = t1.userid
AND t2.userid = t1.matchuserid);

How can I count rows in a 1:N:N relation in a faster way?

This question is a bit complicated to me, and I can't explain it in one sentence so the title may seem quite ambiguous.
I have 3 tables in my MySQL database, their structure is shown below:
word_list (5 million rows)
+-----+--------+
| wid | word |
+-----+--------+
| 1 | foo |
| 2 | bar |
| 3 | hello |
+-----+--------+
paper_word_relation (10 million rows)
+-----+-------+
| pid | word |
+-----+-------+
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 3 |
+-----+-------+
paper_citation_relation (80K rows)
+----------+--------+
| pid_from | pid_to |
+----------+--------+
| 1 | 2 |
| 1 | 3 |
| 1 | 4 |
| 2 | 1 |
| 2 | 3 |
+----------+--------+
I want to find out how many papers contain word W, and cite the papers also contain word W.(for each word in the list)
I use two inner join to do this job but it seems extremely slow when the word is popular - above 50s (quite fast if the word is rarely used - below 0.1s), here is my code
SELECT COUNT(*) FROM (
SELECT a.pid_from, a.pid_to, b.word FROM paper_citation_relation AS a
INNER JOIN paper_word_relation AS b ON a.pid_from = b.pid
INNER JOIN paper_word_relation AS c ON a.pid_to = c.pid
WHERE b.word = 2 AND c.word = 2) AS d
How can I do this faster? Is my query not efficient enough or it's the problem about the amount of data?
I can only come up with one solution that I delete the words which occur less than 2 in the paper_word_relation table. (About 4 million words only occur once)
Thanks!
If you are only concerned with getting the Count, you should not be first getting the results into a Derived Table, and then Count the rows out. This may create unnecessary temporary tables storing lots of data in-memory. You can directly count the number of rows.
I also think that you need to count unique number of papers. Because of Many-to-Many relationships in paper_citation_relation table, duplicate rows may be coming for a single paper.
SELECT COUNT(DISTINCT a.pid_from)
FROM paper_citation_relation AS a
INNER JOIN paper_word_relation AS b ON a.pid_from = b.pid
INNER JOIN paper_word_relation AS c ON a.pid_to = c.pid
WHERE b.word = 2 AND c.word = 2
For performance, you will need following indexing:
Composite Index on (pid_from, pid_to) in the paper_citation_relation table.
Composite Index on (pid, word) in the paper_word_relation table.
We may also possibly optimize the query further by reducing one join and use conditional AND/OR based filtering in HAVING. You will need to benchmark it though.
SELECT COUNT(*)
FROM (
SELECT a.pid_from
FROM paper_citation_relation AS a
INNER JOIN paper_word_relation AS b
ON (a.pid_from = b.pid OR
a.pid_to = b.pid)
GROUP BY a.pid_from
HAVING SUM(a.pid_from = b.pid AND b.word = 2) AND
SUM(a.pid_to = b.pid AND b.word = 2)
)
After the first 1:n join you get the same pid_to multiple times and your next join is no longer 1:n but n:m, creating a possibly huge intermediate result before the final DISTINCT. It's similar to a CROSS JOIN and it's getting worse for popular words, e.g. 10*10 vs. 1000*1000 rows.
You must remove the duplicates before the join, this should return the same number as #MadhurBhaiya's answer
SELECT Count(*) -- no more DISTINCT needed
FROM
(
SELECT DISTINCT cr.pid_to -- reducing m to 1
FROM paper_citation_relation AS cr
JOIN paper_word_relation AS wr
ON cr.pid_from = wr.pid
WHERE wr.word = 2
) AS dt
JOIN paper_word_relation AS wr
ON dt.pid_to = wr.pid -- 1:n join again
WHERE wr.word = 2
If you want to count the number of papers which have been cited you need to get a distinct list of pid (either pid_from or pid_to) from paper_citation_relation first and then join to the specific word.
SELECT Count(*)
FROM
( -- get a unique list of cited or citing papers
SELECT pid_from AS pid -- citing
FROM paper_citation_relation
UNION -- DISTINCT by default
SELECT pid_to -- cited
FROM paper_citation_relation
) AS dt
JOIN paper_word_relation AS wr
ON wr.pid = dt.pid
WHERE wr.word = 2 -- now check for the searched word
The number returned by this might be slightly higher (it counts a paper regardless if cited or citing).

SUM() over a many-to-many relation in MYSQL

I have tables stanje and transakcija in a many-to-many relation, as shown in the image:
I need a MYSQL clause that returns all rows in stanje joined by a SUM() of every transakcija.iznos connected to a given stanje .
So far I have tried
select SUM(t.iznos)
from transakcija t
where transakcija_id in
(select transakcija_id from stanje_transakcija where stanje_id = ?)
which returns the SUM() correctly when given a stanje_id, but have no idea how to proceed, since I need sums for all rows in stanje.
Edit: added example output
------------------------------------
| stanje_id | naziv | SUM(t.iznos) |
------------------------------------
| 1 | a | 125.2 |
| 2 | b | -42.2 |
------------------------------------
If I understand correctly, you need to use JOIN in thoes tables by transakcija_id column and stanje_id column.
From your expect result you can try to use SUM with GROUP BY
select t2.stanje_id,t2.naziv,SUM(t.iznos)
from transakcija t
INNER JOIN stanje_transakcija t1 on t.transakcija_id = t1.transakcija_id
INNER JOIN stanje t2 on t2.stanje_id = t1.stanje_id
GROUP BY t2.stanje_id,t2.naziv

MySQL intersection of two tables

I need to implement a function which returns all the networks the installation is not part of.
Following is my table and for example if my installation id is 1 and I need all the network ids where the installation is not part of then the result will be only [9].
network_id | installation_id
-------------------------------
1 | 1
3 | 1
2 | 1
2 | 2
9 | 2
2 | 3
I know this could be solved with a join query but I'm not sure how to implement it for the same table. This is what I've tried so far.
select * from network_installations where installation_id = 1;
network_id | installation_id
-------------------------------
1 | 1
2 | 1
3 | 1
select * from network_installations where installation_id != 1;
network_id | installation_id
-------------------------------
9 | 2
2 | 2
2 | 3
The intersection of the two tables will result the expected answer, i.e. [9]. But though we have union, intersect is not present in mysql. A solution to find the intersection of the above two queries or a tip to implement it with a single query using join will be much appreciated.
The best way to do this is to use a network table (which I presume exists):
select n.*
from network n
where not exists (select 1
from network_installation ni
where ni.network_id = n.network_id and
ni.installation_id = 1
);
If, somehow, you don't have a network table, you can replace the from clause with:
from (select distinct network_id from network_installation) n
EDIT:
You can do this in a single query with no subqueries, but a join is superfluous. Just use group by:
select ni.network_id
from network_installation ni
group by ni.network_id
having sum(ni.installation_id = 1) = 0;
The having clause counts the number of matches for the given installation for each network id. The = 0 is saying that there are none.
Another solution using OUTER JOIN:
SELECT t1.network_id, t1.installation_id, t2.network_id, t2.installation_id
FROM tab t1 LEFT JOIN tab t2
ON t1.network_id = t2.network_id AND t2.installation_id = 1
WHERE t2.network_id IS NULL
You can check at http://www.sqlfiddle.com/#!9/4798d/2
select *
from network_installations
where network_id in
(select network_id
from network_installations
where installation_id = 1
group by network_id )

How can I get connected values of two times the same foreign key?

I currently have a situation that could easily be solved with 3 SQL queries, but I wonder if it can be done in one query.
I have the following tables:
symbol similarity
------- ------------
id | name | latex id | base_symbol_id | similar_symbol_id
I want to SELECT so that my result looks like this:
query_result
------------
similarity_id | base_formula_id | base_formula_name | base_formula_latex | similar_formula_id | similar_formula_name | similar_formula_latex
Failed tries
I usual solve similar tasks with JOIN. But this time, the SELECT depends on another attribute I select ... I don't know how to do this. Here is my try (which of course failed):
SELECT `base_symbol_id`, `similar_symbol_id`, `latex`
FROM `similarity`
JOIN `symbol` ON ((`symbol`.`id` = `base_symbol_id`) OR (`symbol`.`id` = `similar_symbol_id`))
gives
base_symbol_id | simlar_symbol_id | latex
10 | 11 | \alpha
10 | 11 | a
select sim.id
,base.id
,base.name
,base.latex
,similar.id
,similar.name
,similar.latex
from similarity as sim
join symbol as base on base.id=sim.base_symbol_id
join symbol as similar on similar.id=sim.similar_symbol_id
Using the given table structure, and making up some random sample inputs in an SQL Fiddle session, the following query would work as you desired:
SELECT T.id as similarity_id,
S1.id as base_formula_id, S1.name as base_formula_name, S1.latex as base_formula_latex,
S2.id as similar_formula_id, S2.name as similar_formula_name, S2.latex as similar_formula_latex
FROM similarity T
LEFT OUTER JOIN symbol S1 ON (T.base_symbol_id = S1.id)
LEFT OUTER JOIN symbol S2 ON (T.similar_symbol_id = S2.id)