I have 1 table with the following cols:
giver_id | receiver_id
10 | 12
9 | 10
10 | 20
12 | 10
I am looking for a mysql query that will return 10-12 / 12-10 as a match.
Thanks
To identify the records for which an "opposite" record exist, you could do:
SELECT *
FROM mytable t
WHERE EXISTS (
SELECT 1
FROM mytable t1
WHERE t1.giver_id = t.receiver_id AND t.giver_id = t1.receiver_id
)
This demo on DB Fiddle with your sample data returns:
| giver_id | receiver_id |
| -------- | ----------- |
| 10 | 12 |
| 12 | 10 |
Related
I have customer table with 10 columns. In the table customer id is repeated. I need to take only one record every customer but randomly.
Let suppose customer table contain total 10000 records. But distinct customers is only 500.
So i need only 500 distinct customer data randomly.
I am using mysql 5.7.
Consider the following...
SELECT * FROM my_table;
+----+-------------+
| id | customer_id |
+----+-------------+
| 1 | 1 |
| 2 | 1 |
| 3 | 3 |
| 4 | 5 |
| 5 | 3 |
| 6 | 2 |
| 7 | 1 |
| 8 | 4 |
| 9 | 5 |
| 10 | 2 |
| 11 | 3 |
| 12 | 1 |
| 13 | 4 |
+----+-------------+
SELECT id
, customer_id
FROM
( SELECT id
, customer_id
, CASE WHEN #prev=customer_id THEN #i:=#i+1 ELSE #i:=1 END i
, #prev:=customer_id
FROM
( SELECT id
, customer_id
FROM my_table
ORDER
BY customer_id
, RAND()
) x
JOIN (SELECT #prev:=null,#i:=0) vars
) n
WHERE i = 1
ORDER
BY customer_id;
-- sample output, different each time --
+----+-------------+
| id | customer_id |
+----+-------------+
| 12 | 1 |
| 10 | 2 |
| 3 | 3 |
| 8 | 4 |
| 9 | 5 |
+----+-------------+
You do not want to ORDER BY RAND() because that will be extremely slow for a large table because it will actually sort all of those random records.
Instead pick a random int less than the number of rows in the table (random_num_less_than_row_count) and do this which is faster but not perfect.
SELECT * FROM atable LIMIT $random_num_less_than_row_count, 1
Or if u have a primary key that is an auto_increment you can pick a random int less than the highest id in the table (random_num_less_than_last_id) do the following which is pretty fast.
SELECT * FROM atable WHERE id >= $random_num_less_than_last_id ORDER BY id ASC LIMIT 1
I did a >= and an ORDER BY id ASC so that if you are missing ids you'll still get a result. But if you have many large gaps you need the slower first option above.
Not sure about it but it is a beginner level query which might to get the desired result
SELECT Distinct column FROM table
ORDER BY RAND()
LIMIT 500
PS: This code isn't in mysql 5.7. And if anyone have a better query more than happy to get corrected
I have a problem with SQLcode
I have a table
id | content | id_user | id_store
1 | abc | 1 | 10
2 | xzy | 1 | 10
3 | abc | 1 | 10
4 | abc | 1 | 11
5 | abc | 1 | 12
My problem is how i got the result is the count of max (id_store) which is 2* value >= max(id_store)
This is a example, result will be
id_store | count(...)
10 | 3
because (3*2) > max of count = 3
Tks everyone
It's very difficult to understand your question. Try to use the next query
SELECT id_store,COUNT(*) CountOfStore
FROM `Your Table`
GROUP BY id_store
HAVING 2*COUNT(*) >= (
SELECT MAX(CountOfStore) -- max of all CountOfStore
FROM
(
SELECT COUNT(*) CountOfStore -- count of store for each id_store
FROM `Your Table`
GROUP BY id_store
)
)
Hope I understood you rightly.
I have a table like this:
| id | date | user_id |
----------------------------------------------------
| 1 | 2008-01-01 | 10 |
| 2 | 2009-03-20 | 15 |
| 3 | 2008-06-11 | 10 |
| 4 | 2009-01-21 | 15 |
| 5 | 2010-01-01 | 10 |
| 6 | 2011-06-01 | 10 |
| 7 | 2012-01-01 | 10 |
| 8 | 2008-05-01 | 15 |
I’m looking for a solution how to select user_id where the difference between MIN and MAX dates is more than 3 yrs. For the above data I should get:
| user_id |
-----------------------
| 10 |
Anyone can help?
SELECT user_id
FROM mytable
GROUP BY user_id
HAVING MAX(`date`) > (MIN(`date`) + INTERVAL '3' YEAR);
Tested here: http://sqlize.com/MC0618Yg58
Similar to bernie's approach, I'd keep date formats native. I'd also probably list the MAX first as to avoid an ABS call (secure a positive number is always returned).
SELECT user_id
FROM my_table
WHERE DATEDIFF(MAX(date),MIN(date)) > 365
DATEDIFF just returns delta (in days) between two given date fields.
SELECT user_id
FROM (SELECT user_id, MIN(date) m0, MAX(date) m1
FROM table
GROUP by user_id)
HAVING EXTRACT(YEAR FROM m1) - EXTRACT(YEAR FROM m0) > 3
SELECT A.USER_ID FROM TABLE AS A
JOIN TABLE AS B
ON A.USER_ID = B.USER_ID
WHERE DATEDIFF(A.DATE,B.DATE) > 365
I have one problem that I can't resolve.
I have 2 tables.
Table 1:
ID | Time
1 | 08:12:54
2 | 08:15:40
3 | 09:30:01
4 | 10:15:15
5 | 10:56:12
6 | 11:00:03
Table 2:
ID | Name| Previous | Current
1 | Queue | null | 11
2 | Queue | 11 | 19
3 | Queue | 19 | 11
3 | List | null | 11
4 | Queue | 11 | 16
4 | List | null | 11
5 | Queue | null | 15
6 | Queue | 15 | 19
The result wanted:
NumberQueue | Start | End
11 | 08:12:54 | 08:15:40
19 | 08:15:40 | 09:30:01
11 | 09:30:01 | 10:15:15
15 | 10:56:12 | 11:00:03
...
...
The previous and the current fields, have the number of the Queue and I want to know for each Queue, the start date and the end date, knowing that the previous has the previous Queue, and the current has the new Queue.
I want one query that can present this result. Help me. :(
Regards.
SELECT t1outer.ID, t1outer.Time AS start, (
SELECT Time FROM Table1 AS t1inner
WHERE t1inner.ID > t1outer.ID
ORDER BY ID ASC LIMIT 1
) AS end, Table2.Previous, Table2.Current
FROM Table1 AS t1outer
LEFT JOIN Table2 USING (ID);
This select statement should provide the information you need:
SELECT Current AS Number, t1out.Time AS Start, (
SELECT Time FROM Table1 AS t1in
WHERE t1in.ID > t1out.ID
ORDER BY ID ASC LIMIT 1
) AS End FROM Table2
LEFT JOIN Table1 AS t1out USING (ID)
WHERE Table2.Name = 'Queue';
I have the next example table:
id | user_id | data
-------------------
1 | 1 | 10
2 | 2 | 10
3 | 2 | 10
4 | 1 | 10
5 | 3 | 10
6 | 4 | 10
7 | 4 | 10
8 | 5 | 10
9 | 5 | 10
10 | 2 | 10
11 | 6 | 10
12 | 3 | 10
13 | 1 | 10
I need to create a SELECT query, that LIMITS my data. For example, I have a limit range (1, 3) (page number = 1, row count = 3). It should selects rows with first 3 unique user_id. And if there are some rows in the end of table with this first user_id's, they should be included to the result. LIMIT statement is bad for this query, because I can get more than 3 rows. Output for my limit should be:
id | user_id | data
-------------------
1 | 1 | 10
2 | 2 | 10
3 | 2 | 10
4 | 1 | 10
5 | 3 | 10
10 | 2 | 10
12 | 3 | 10
13 | 1 | 10
Can you help me to generate this query?
How about:
SELECT *
FROM table
WHERE user_id IN
(SELECT distinct(user_id) FROM table order by user_id LIMIT 3);
What about something like this?
SELECT * FROM table WHERE user_id BETWEEN (number) AND (number+row count)
I know it isn't working but you should be able to make it work ^^
The sample code below can be used for Oracle & Mysql. (use TOP for SQL Server & Sybase)
You get all the results from your table (t1) that match the top 3 user_id (t2) (check the MySQL manual for the limit function)
SELECT *
FROM exampletable t1
INNER JOIN (
SELECT DISTINCT user_id
FROM exampletable
ORDER BY user_id
LIMIT 0,3 -- this is the important part
) AS t2 ON t1.user_id = t2.user_id
ORDER BY id
For the next 3 id's change the limit 0,3 to limit 3,6.