I the following table:
ID TIMESLOT_ID SLOTS_AVAILABLE
1 1 10
2 3 2
3 8 3
4 9 10
5 2 10
6 6 10
7 4 10
I want to perform a query where I can get all rows where SLOTS_AVAILABLE is equal or greater than (user_input_1) AND the next (user_input_2) consecutive rows following to that one meet the same condition (SLOTS_AVAILABLE is equal or greater than (user_input_1))
So if (user_input_1) is 3 and (user_input_2) is 3 the result would be:
ID TIMESLOT_ID SLOTS_AVAILABLE
3 8 3
4 9 10
Hmmm . . . If I understand correctly, you want window functions:
select t.*
from (select t.*,
sum(case when slots_available >= :user_input_1 then 1 else 0 end) over
(order by id
rows between current row and :user_input_2 following
) as cnt
from t
) t
where cnt = :user_input_2 + 1;
This requires MySQL 8+.
Here is a db<>fiddle.
Related
I have a table that looks like this
id
1
2
4
5
6
10
11
So a bunch of consecutive values, an unknown number of absent fields and then other consecutive values.
What I am trying to achieve is to get
id
stint
1
0
2
0
4
1
5
1
6
1
10
2
11
2
By incrementing every time the number of the stint, which I can later use for summing over other columns.
Is it possible? Thanks
If your MySQL version support window function.
You can try to use LAG window function in subquery to get previous id column, then use SUM condition aggregate window function.
Query #1
SELECT Id,
SUM(id - n_Id > 1) OVER(ORDER BY id) stint
FROM (
SELECT *,LAG(id,1,id) OVER(ORDER BY id) n_Id
FROM T
) t1
Id
stint
1
0
2
0
4
1
5
1
6
1
10
2
11
2
View on DB Fiddle
I currently have this set of data.
Id
pattern
1
1
2
2
3
1
4
2
5
3
6
4
7
1
8
2
9
1
10
2
11
3
I have ids sorted ascending, and a column called pattern. The goal is to retrieve only the rows with a consecutive pattern >=3 down the table.
So the result would be:
Id
pattern
3
1
4
2
5
3
6
4
9
1
10
2
11
3
Since the pattern for Ids 3-6 have consecutively reached >=3, we will return these 4 rows.
Same goes for Ids 9-11.
Id 1-2 did not reach the threshold of 3. Same for Id 7-8.
Would appreciate any help!
Use SUM() window function to create the groups of the sequences and with MAX() window function check the max pattern in each group:
SELECT Id, pattern
FROM (
SELECT *, MAX(pattern) OVER (PARTITION BY grp) max_pattern
FROM (
SELECT *, SUM(pattern = 1) OVER (ORDER BY Id) grp
FROM tablename
) t
) t
WHERE max_pattern >= 3;
Or, with a correlated subquery, for versions of MySql that do not support window functions:
SELECT t1.*
FROM tablename t1
WHERE (SELECT t2.pattern FROM tablename t2 WHERE t2.Id = t1.Id + (3 - t1.pattern)) = 3;
See the demo.
This question already has an answer here:
How to write these complex MySQL queries?
(1 answer)
Closed 1 year ago.
I am a beginner at databases. I need to write some SQL queries.
The tables are:
Expedition(id, number, id_captain, id_ship, id_heros)
City(id, name)
Heros(id, family_name, first_name)
Step(id, index, id_expedition, id_city)
sample data :
'Table expedition'
id
number
id_captain
id_ship
id_hero
1
1
1
10
8
2
2
2
1
5
3
3
1
8
3
4
4
10
9
6
5
5
5
7
4
6
6
6
5
4
7
7
7
3
7
8
8
8
2
8
9
9
9
1
3
10
10
1
4
2
11
11
6
3
1
12
12
8
6
1
13
13
5
8
6
14
14
4
9
9
15
15
3
10
4
16
16
10
2
2
17
17
9
3
3
18
18
8
7
7
19
19
9
8
10
20
20
7
2
2
table 'heros'
id
family_name
first_name
1
familyname1
firstname1
2
familyname2
firstname2
3
familyname3
firstname3
4
familyname4
firstname4
5
familyname5
firstname5
6
familyname6
firstname6
7
familyname7
firstname7
8
familyname8
firstname8
9
familyname9
firstname9
10
familyname10
firstname10
query1: The family (based on the family name) with the least travelling (the fewest cities different crossings).
i have done this for the first query:
select expedition.id, id_hero, heros.family_name as Famille_expedition, count(distinct id_city) as city_count
from expedition, step, heros
where expedition.id=step.id and expedition.id_hero=heros.id
group by id_hero
having city_count =
(select count(distinct id_city) as min_city_count
from expedition, step
where expedition.id=step.id
group by id_hero
order by min_city_count asc
limit 1);
query2: The average of cities crossed by an expedition
I have no idea how to answer the second one.
Well, first ask yourself what information do you need to answer your question?
From your question, I'd say the average number of crossings is just the sum of all entries in the steps table, divided by the number of expeditions, since in each step, one city is visited and the average of all visits is what you are looking for:
SELECT (
(SELECT COUNT(s.id_city)
FROM step AS s) /
(SELECT COUNT(e.id)
FROM expedition AS e) ) AS total_average__cities
That being said, it depends on how exactly you define number of cities and crossing. Imagine the following example data for the table step:
id
idx
id_expedition
id_city
1
1
1
1
2
2
1
5
3
3
1
3
4
1
2
5
5
2
2
9
6
1
3
8
7
2
3
5
8
3
3
9
9
4
3
5
10
5
3
8
The table lists the steps for three expeditions. Expedition 1 goes from one city via another to a third. Expedition 2 goes directly from one city to another. And expedition 3 goes through several cities and visits one city twice along the way and also returns to the city that it started in.
The average number of cities over all these steps is (3 + 2 + 5 [cities in all steps]) / 3 [expeditions] = 3.3333. That is the result of the above query.
Now, if you define number of cities as meaning unique cities for each expedition, expedition 3 only visits 3 cities instead of 5. Then your average calculates as (3 + 2 + 3 [unique cities/expedition in all steps]) / 3 [expeditions] = 2.6666. The according query needs to count the distinct cities within each expedition before building the average:
SELECT (
(SELECT SUM(cnt) FROM (SELECT COUNT(DISTINCT s.id_city) AS cnt
FROM step AS s
GROUP BY s.id_expedition) t) /
(SELECT COUNT(e.id)
FROM expedition AS e) ) AS total_average__cities
Now, if you define crossing as only covering cities along the way, expedition 1 only crosses 1 city and expedition 2 crosses none at all.
Then your query also needs to look differently. You need to filter the all cities to exclude the first and the last for each expedition. The subquery could look like this:
SELECT s.* FROM step s
JOIN ( SELECT id_expedition,
MAX(idx) AS max_idx,
MIN(idx) AS min_idx
FROM step s
GROUP BY id_expedition) minmax
ON s.id_expedition = minmax.id_expedition
AND s.idx > minmax.min_idx
AND s.idx < minmax.max_idx
So for the case that you want the number of cities crossed excluding start and stop, your average would be computed as (1 + 0 + 3 [intermediate cities in all steps]) / 3 [expeditions] = 1.3333. The according query would be
SELECT (
(SELECT COUNT(s.id_city)
FROM step s
JOIN ( SELECT id_expedition,
MAX(idx) as max_idx,
MIN(idx) as min_idx
FROM step s
GROUP BY id_expedition) minmax
ON s.id_expedition = minmax.id_expedition
AND s.idx > minmax.min_idx
AND s.idx < minmax.max_idx) /
(SELECT COUNT(e.id)
FROM expedition AS e) ) AS total_average__cities
Finally, in case you want to both exclude start and stop and only want to count unique cities, your average would be computed as (1 + 0 + 2 [unique intermediate cities in all steps]) / 3 [expeditions] = 1. The following query combines the two approaches from above:
SELECT (
(SELECT SUM(cnt) FROM (SELECT COUNT(DISTINCT id_city) AS cnt
FROM step s
JOIN ( SELECT id_expedition,
MAX(idx) AS max_idx,
MIN(idx) AS min_idx
FROM step s
GROUP BY id_expedition) minmax
ON s.id_expedition = minmax.id_expedition
AND s.idx > minmax.min_idx
AND s.idx < minmax.max_idx
GROUP BY s.id_expedition) t) /
(SELECT COUNT(e.id)
FROM expedition AS e) ) AS total_average_cities
You can test all these queries in this db<>fiddle.
I have a database with a table called BOOKINGS containing the following values
main-id place-id start-date end-date
1 1 2018-8-1 2018-8-8
2 2 2018-6-6 2018-6-9
3 3 2018-5-5 2018-5-8
4 4 2018-4-4 2018-4-5
5 5 2018-3-3 2018-3-10
5 1 2018-1-1 2018-1-6
4 2 2018-2-1 2018-2-10
3 3 2018-3-1 2018-3-28
2 4 2018-4-1 2018-4-6
1 5 2018-5-1 2018-5-15
1 3 2018-6-1 2018-8-8
1 4 2018-7-1 2018-7-6
1 1 2018-8-1 2018-8-18
1 2 2018-9-1 2018-9-3
1 5 2018-10-1 2018-10-6
2 5 2018-11-1 2018-11-5
2 3 2018-12-1 2018-12-25
2 2 2018-2-2 2018-2-19
2 4 2018-4-4 2018-4-9
2 1 2018-5-5 2018-5-23
What I need to do is for each main-id I need to find the largest total number of days for every place-id. Basically, I need to determine where each main-id has spend the most time.
This information must then be put into a view, so unfortunately I can't use temporary tables.
The query that gets me the closest is
CREATE VIEW `MOSTTIME` (`main-id`,`place-id`,`total`) AS
SELECT `BOOKINGS`.`main-id`, `BOOKINGS`.`place-id`, SUM(DATEDIFF(`end-date`, `begin-date`)) AS `total`
FROM `BOOKINGS`
GROUP BY `BOOKINGS`.`main-id`,`RESERVATION`.`place-id`
Which yields:
main-id place-id total
1 1 24
1 2 18
1 5 5
2 1 2
2 2 20
2 4 9
3 1 68
3 2 24
3 3 30
4 1 5
4 2 10
4 4 1
5 1 19
5 2 4
5 5 7
What I need is then the max total for each distinct main-id:
main-id place-id total
1 1 24
2 2 20
3 1 68
4 2 10
5 1 19
I've dug through a large amount of similar posts that recommend things like self joins; however, due to the fact that I have to create the new field total using an aggregate function (SUM) and another function (DATEDIFF) rather than just querying an existing field, my attempts at implementing those solutions have been unsuccessful.
I am hoping that my query that got me close will only require a small modification to get the correct solution.
Having hyphen character - in column name (which is also minus operator) is a really bad idea. Do consider replacing it with underscore character _.
One possible way is to use Derived Tables. One Derived Table is used to determine the total on a group of main id and place id. Another Derived Table is used to get maximum value out of them based on main id. We can then join back to get only the row corresponding to the maximum value.
CREATE VIEW `MOSTTIME` (`main-id`,`place-id`,`total`) AS
SELECT b1.main_id, b1.place_id, b1.total
FROM
(
SELECT `main-id` AS main_id,
`place-id` AS place_id,
SUM(DATEDIFF(`end-date`, `begin-date`)) AS total
FROM BOOKINGS
GROUP BY main_id, place_id
) AS b1
JOIN
(
SELECT dt.main_id, MAX(dt.total) AS max_total
FROM
(
SELECT `main-id` AS main_id,
`place-id` AS place_id,
SUM(DATEDIFF(`end-date`, `begin-date`)) AS total
FROM BOOKINGS
GROUP BY main_id, place_id
) AS dt
GROUP BY dt.main_id
) AS b2
ON b1.main_id = b2.main_id AND
b1.total = b2.max_total
MySQL 8+ solution would be utilizing the Row_Number() functionality:
CREATE VIEW `MOSTTIME` (`main-id`,`place-id`,`total`) AS
SELECT b.main_id, b.place_id, b.total
FROM
(
SELECT dt.main_id,
dt.place_id,
dt.total
ROW_NUMBER() OVER (PARTITION BY dt.main_id
ORDER BY dt.total DESC) AS row_num
FROM
(
SELECT `main-id` AS main_id,
`place-id` AS place_id,
SUM(DATEDIFF(`end-date`, `begin-date`)) AS total
FROM BOOKINGS
GROUP BY main_id, place_id
) AS dt
GROUP BY dt.main_id
) AS b
WHERE b.row_num = 1
i have a problem concerning a select query in MYSQL
i have two different tables and i want to obtain a certain result
i used COUNT method which gave me only the results (>=1)
But in reality , i want to use all counts with zero included how to do it?
My query is:
SELECT
first.subscriber_id,
second.tag_id,
COUNT(*)
FROM
content_hits first
JOIN content_tag second ON first.content_id=second.content_id
GROUP BY
second.Tag_id,first.Subscriber_id<br>
First table:Content_hits
CONTENT_ID SUBSCRIBER_ID
30 1
10 10
34 4
32 2
40 3
28 3
30 6
31 8
12 3
Second table:Content_tag
CONTENT_ID TAG_ID
1 1
2 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
11 2
12 2
13 2
14 2
Result but incomplete For example:Subsrciber6 for tag_id=1 should have a count(*)=0
subscriber_id tag_id COUNT(*)
1 1 4
2 1 7
3 1 2
4 1 1
5 1 3
7 1 2
8 1 1
9 1 1
10 1 3
1 2 2
2 2 3
3 2 2
Now that you have further elaborated on what you actually want to achieve, it can be seen that the problem is much more complex. You actually want all combinations of subscriber_id and tag_id, and then count the number of actual entries in the joined table product. whew. So here goes the SQL:
SELECT combinations.tag_id,
combinations.subscriber_id,
-- correlated subquery to count the actual hits by tag/subscriber when joining
-- the two tables using content_id
(SELECT count(*)
FROM content_hits AS h
JOIN content_tag AS t ON h.content_id = t.content_id
WHERE h.subscriber_id = combinations.subscriber_id
AND t.tag_id = combinations.tag_id) as cnt
-- Create all combinations of tag/subscribers first, before counting anything
-- This will be necessary to have "zero-counts" for any combination of
-- tag/subscriber
FROM (
SELECT DISTINCT tag_id, subscriber_id
FROM content_tag
CROSS JOIN content_hits
) AS combinations
Not sure, but is this what you want?
SELECT first.subscriber_id, second.tag_id, COUNT(*) AS c
FROM content_hits first JOIN content_tag second ON first.content_id=second.content_id
GROUP BY second.Tag_id,first.Subscriber_id HAVING c = 0