Need MySQL query for below problem
Consider a table having student and their marks in a particular subject
Schema
std_id int(11)
marks int(11)
Sample data
std_id marks
1 10
2 15
3 90
4 120
5 25
6 29
7 121
8 122
Now I have an web app in which a form will take a input (int) from user.
For eg 12
then I am required to show total number of student ids (std_id) and their corresponding marks group.
Eg
std_total (tot no of students) group (marks range we got from form)
1 0-11
1 12-23
2 24-35
1 84-95
3 120-131
#Barmar Your answer was almost correct, I made few changes to clean the output. Your query gives output as below :
0-11 2
1-12 2
2-13 1
3-14 1
4-15 1
6-17 1
7-18 2
My query return Outout as
0-11 2
12-23 2
24-35 1
36-47 1
48-59 1
72-83 1
84-95 2
SELECT CONCAT(FLOOR(marks/12)*12, '-', FLOOR(marks/12)+11*(FLOOR(marks/12))+11) AS `group`, COUNT(*) as `std_total`
FROM yourTable
GROUP BY `group`
Use division and FLOOR() to get the beginning of each range.
SELECT CONCAT(FLOOR(marks/12), '-', FLOOR(marks/12)+11) AS `group`, COUNT(*) as `std_total`
FROM yourTable
GROUP BY `group`
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 have a table where each quiz ID is repeated multiple times. there is a date in front of each quiz id in each row. I want to select entire row for each quiz ID where date is latest with user. The date format is mm/dd/YYYY.
Sample -
USER_ID Quiz_id Name Date Marks .. .. ..
1 2 poly 4/3/2020 27
1 2 poly 4/3/2019 98
1 4 moro 4/3/2020 09
2 5 cat 4/12/2015 87
2 4 moro 4/3/2009 56
2 6 PP 4/3/2011 76
3 2 poly 4/3/2020 12
3 2 poly 5/3/2020 09
3 7 dog 4/3/2011 23
I want result look like this:Result
USER_ID Quiz_id Name Date Marks .. .. ..
1 2 poly 4/3/2020 27
1 4 moro 4/3/2020 09
2 5 cat 4/12/2015 87
2 4 moro 4/3/2009 56
2 6 PP 4/3/2011 76
3 2 poly 5/3/2020 09
3 7 dog 4/3/2011 23
You can use rank function to get the desired result:
Demo
SELECT A.* FROM (
SELECT A.*, RANK() OVER(PARTITION BY USER_ID,QUIZ_ID, NAME ORDER BY DATE DESC) RN FROM
Table1 A ORDER BY USER_ID) A WHERE RN = 1 ORDER BY USER_ID, QUIZ_ID;
I don't have MySQL installed so you will need to test and report back. The general idea is to identify the row of interest using max and a group by (table t). As the Date column appears to be text column (MySQL uses the format YYYY-MM-DD for dates) you will need to convert it to a date with str_to_date() so you can use the max() aggregate function. Finally, join with the original table (here table t2 to do the date conversion), as only the aggregate column(s) and columns named in the group by are well defined (in table t1), i.e.:
select USER_ID, Quiz_id, Date, Marks from (
select USER_ID, Quiz_id, max(str_to_date(Date, '%m/%d/%Y')) as Date2 from quiz group by 1, 2
) as t natural join (
select *, str_to_date(Date, '%m/%d/%Y') Date2 from Quiz
) as t2;
I don't recall off-hand but Date might be reserved word, in which case you will need to quote the column name, or ideally rename said column to use a better name.
Also, the original table is not in 3rd normal form as Quiz_id depends on Name. Quiz_id, as implied, should be a foreign key to a lookup table that holds the Name.
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 my data base like this
id project_id client_id price
1 1 1 200
2 2 1 123
3 2 1 100
4 1 1 87
5 1 1 143
6 1 1 100
7 3 3 123
8 3 3 99
9 4 3 86
10 4 3 43
11 4 3 145
12 4 3 155
Now here I want that it will sum the price columns with the same client_id.
For that I just made my query like this
Select `project_id`, SUM(`price`) FROM `table-name` GROUP BY `client_id`
This one is doing sum the price but I am getting only two project_id in the result. I want the result should be all the distinct project for the client id and the price will be summed for the group clients.
So can someone tell me how to do this? Any help and suggestions will be really appreciable. Thanks
You should not have "bare" column in a group by query that are not in the group by statement.
If you want the list of projects, you can get them in a list like this:
SELECT client_id, GROUP_CONCAT(project_id), SUM(price)
FROM table-name
GROUP BY client_id;
you only have two client that why you are getting only two record , you can group by two column,
Select `project_id`, SUM(`price`) FROM `table-name` GROUP BY `client_id`, `project_id`
I want to find number of repeated value in group using single select query.
I can't use the column in group by for which I want to find number of occurrence in some group.
e.g.
obsid Name Value
1 Ronak 120
2 Ronak 125
3 Ronak 120
4 Pankti 130
5 Pankti 130
6 Sanket 140
7 Aakash 140
8 Unnat 120
Now I want to develop select query in mysql that give me below result
obsid Name Value Value_occurrence
1 Ronak 120 2
2 Ronak 125 1
4 Pankti 130 2
6 Sanket 140 1
7 Aakash 140 1
8 Unnat 120 1
SELECT min(obsid) AS obsid, Name, Value, COUNT(*) AS Value_occurence
FROM yourTable
GROUP BY Name, Value
Group by the desired columns and use COUNT to count the rows in each group:
SELECT MIN(`obsid`) AS `obsid`, `Name`, `Value`, COUNT(*) AS Value_occurrence
FROM yourtable
GROUP BY `Name`, `Value`
Try this online demo which shows your example data and the results for the above query.