I've a table looking more or less like that:
**Day** | **Mileage**
----------------
1 | 13
2 | 2
3 | 25
4 | 15
5 | 20
6 | 8
7 | 17
8 | 12
9 | 16
10 | 5
How to write a SQL query:
Returning total mileage from the firts to the nth day without using php? For example, day 1:13, day 2: 15, day 3: 40.
How to get 5 biggest mileages without using limit?
For the sum up:
select mday, (select SUM(miles) from mm s where s.mday <= mm.mday) tot_miles
from mm;
For the 5 biggest mileage per day
select mday, miles from mm order by miles desc limit 5
Change the column names and table name to match yours.
Since you added the "no use of limit", which is odd, you can try this:
set #r =0;
select * from (
select *, #r :=#r+1 as r from mm order by miles desc) e
where r<=5
Since You do not want to use limit here is the tedious way
1)
select sum(mileage) from xyz where day between 2 and 15;
2)
mysql> select #rownum:=#rownum+1 'rank',mileage
from (select mileage from xyzorder by mileage desc)mil,(select #rownum:=0)r
where #rownum<5;
Related
I have 3 columns, employee_id, start_time and end_time I want to make bucks of 1 hour to show me how many employees were working in each hour. For example, employee A worked from 12 pm to 3 pm and employee B worked from 2 pm to 4 pm so, at 12 pm (1 employee was working) 1 pm (1 employee) 2 pm (2 employees were working) 3 pm (2 employees) and 4 pm (1 employee), how can I make this in SQL? Let me show you a picture of the start and end time columns.
Sample input would be:
Expected outcome would be something like
I want to create a bucket in order to know how many people were working in each hour of the day.
SELECT
Employee_id,
TIME(shift_start_at,timezone) AS shift_start,
TIME(shift_end_at,timezone) AS shift_end,
FROM
`employee_shifts` AS shifts
WHERE
DATE(shifts.shift_start_at_local) >= "2022-05-01"
GROUP BY
1,
2,
3
Assuming you are on mysql version 8 or above generate all the buckets , left join to shifts to infill times in start-endtime ranges , filter out those that are not applicable then count eg:-
DROP TABLE IF EXISTS t;
create table t (id int, startts datetime, endts datetime);
insert into t values
(1,'2022-06-19 08:30:00','2022-06-19 10:00:00'),
(2,'2022-06-19 08:30:00','2022-06-19 08:45:00'),
(3,'2022-06-19 07:00:00','2022-06-19 07:59:00');
with cte as
(select 7 as bucket union select 8 union select 9 union select 10 union select 11),
cte1 as
(select bucket,t.*,
floor(hour(startts)) starthour, floor(hour(endts)) endhour
from cte
left join t on cte.bucket between floor(hour(startts)) and floor(hour(endts))
)
select bucket,count(id) nof from cte1 group by bucket
;
+--------+-----+
| bucket | nof |
+--------+-----+
| 7 | 1 |
| 8 | 2 |
| 9 | 1 |
| 10 | 1 |
| 11 | 0 |
+--------+-----+
5 rows in set (0.001 sec)
If you have a limited number of time bucket maybe you can use it this way
WITH CTE AS
(SELECT
COUNTRY,
MONTH,
TIMESTAMP_DIFF(time_b, time_a, MINUTE) dt,
METRIC_a,
METRIC_b
FROM
TABLE_NAME)
SELECT
CASE
WHEN dt BETWEEN 0 AND 10 THEN "0-10"
WHEN dt BETWEEN 10 AND 20 THEN "11-20"
WHEN dt BETWEEN 20 AND 30 THEN "21-30"
WHEN dt BETWEEN 30 AND 40 THEN "31-40"
WHEN dt > 40 THEN ">40"
END as time_bucket,
AVG(METRIC_a),
SUM(METRIC_b)
FROM CTE
Althought, I should emphasize that this solution works if you have a limited bucket. If you have a lot of buckets, you can create a base table with your buckets then LEFT JOIN it to get your results.
Just use a subquery for each column mentioning the required timestamp in between, also make sure your start_time and end_time columns are timestamp types. For more information, please share the table structure, sample data, and expected output
If I understood well, this would be
SELECT HOUR, (SELECT COUNT(*)
FROM employee
WHERE start_time <= HOUR
AND end_time >= HOUR) AS working
FROM schedule HOUR
Where schedule is a table with employee schedules.
I have data of an event with duration (say, eating a meal at a restaurant) and I want to know for any given hour how many events were taking place. The data looks like this:
Event | Start Time | End Time
-----------------------------------------
1 | 12:03 | 14:20
2 | 12:30 | 12:50
3 | 13:05 | 14:45
4 | 14:01 | 14:49
I also have "Duration" available as an alternative to "End Time". The result I'm looking for would be like this:
Hour | Count
-----------------------
12 | 2
13 | 2
14 | 3
During hour 12, there were two events happening (1 & 2), hour 13 also had two events (1 & 3) and hour 14 had three events (1, 3, & 4).
I can do this programmatically with a loop. I can count when the events start (or end) in SQL. But I'd really like to bridge the gap and do this in SQL, but I can't think of a way.
One possible solution (works with MySQL v5.6+ and SQLite3):
create table hours(Hour int);
insert into hours values
(0),(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12),
(13),(14),(15),(16),(17),(18),(19),(20),(21),(22),(23);
create table log(Event int,StartTime varchar(5),EndTime varchar(5));
insert into log values
(1,'12:03','14:20'),
(2,'12:30','12:50'),
(3,'13:05','14:45'),
(4,'14:01','14:49');
-- ------------------------------------------------------------------------------
select Hour,count(Event) Count
from log join hours
on Hour between substr(StartTime,1,2) and substr(EndTime,1,2)
group by Hour;
If you are running MySQL 8.0, you could use UNION ALL, window functions and aggregation, like so:
select hr, sum(sum(cnt)) over(order by hr) cnt
from (
select hour(start_time) hr, 1 cnt from mytable
union all select hour(end_time) + 1, -1 from mytable
) t
group by hr
Demo on DB Fiddle:
hr | cnt
-: | --:
12 | 2
13 | 2
14 | 3
15 | 0
If you do not have MySql 8, then create a table hour:
CREATE TABLE hour (
hr INT PRIMARY KEY
);
INSERT INTO hour(hr) VALUES
(0),(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),
(12),(13),(14),(15),(16),(17),(18),(19),(20),(21),(22),(23);
And then:
select h.hr, count(*) as cnt from hour h
join mytable m on h.hr between hour(m.Start_Time) and hour(m.End_Time)
group by hr
order by hr
;
See 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 database containing information about sick days of employees in the following structure ( example ):
date || login
2018-01-02 || TestLogin1
2018-01-03 || TestLogin2
2018-01-04 || TestLogin5
2018-01-05 || TestLogin1
2018-01-06 || TestLogin2
And I want to check whether someone had 23 Sick Days in a row within previous 60 days.
I know how to do this in PHP, using loops , but was wondering whether there is a possibility to create this app in raw MySQL.
This is the output I want to achieve:
login || NumberOfDaysOnSickLeaveWithinPrevious2Month
TestLogin4 || 32
TestLogin7 || 30
TestLogin12 || 20
TestLogin3 || 15
TestLogin1 || 10
Will be thankful for the support,
Thanks in advance,
Your sample data suggests that you just want aggregation:
select login,
count(*) as NumberOfDaysOnSickLeaveWithinPrevious2Month
from t
where date >= curdate() - interval 2 month
group by login;
That has nothing to do with "consecutive days". But your sample data doesn't even show two days in a row with the same login -- nor even any dates within the past two months.
It's a lot easier to develop this if you shrink the numbers for example 2 or more continuous days absent in the last 5 days.
drop table if exists t;
create table t(employee_id int, dt date);
insert into t values
(1,'2018-07-10'),(1,'2018-07-11'),(1,'2018-07-12'),
(2,'2018-07-10'),(2,'2018-07-15'),
(3,'2018-07-10'),(3,'2018-07-11'),(3,'2018-07-13'),(3,'2018-07-14')
;
select employee_id, bn, count(*)
from
(
select t.*, concat(employee_id,year(dt) * 10000 + month(dt) * 100 + day(dt))
- #p = 1 diff,
if(
concat(employee_id,year(dt) * 10000 + month(dt) * 100 + day(dt))
- #p = 1 ,#bn:=#bn,#bn:=#bn+1) bn,
#p:=concat(employee_id,year(dt) * 10000 + month(dt) * 100 + day(dt)) p
from t
cross join (select #bn:=0,#p:=0) b
where dt >= date_add(date(now()), interval -5 day)
order by employee_id,dt
) s
group by employee_id,bn having count(*) >= 2 ;
+-------------+------+----------+
| employee_id | bn | count(*) |
+-------------+------+----------+
| 1 | 1 | 3 |
| 3 | 4 | 2 |
| 3 | 5 | 2 |
+-------------+------+----------+
3 rows in set (0.06 sec)
Note the use of variables to work out a block number ,and the having clause. Concating employee and date creates a psuedo key and simplifies calculation.
I am having an issue with a SELECT command in MySQL. I have a database of securities exchanged daily with maturity from 1 to 1000 days (>1 mio rows). I would like to get the outstanding amount per day (and possibly per category). To give an example, suppose this is my initial dataset:
DATE VALUE MATURITY
1 10 3
1 15 2
2 10 1
3 5 1
I would like to get the following output
DATE OUTSTANDING_AMOUNT
1 25
2 35
3 15
Outstanding amount is calculated as the total of securities exchanged still 'alive'. That means, in day 2 there is a new exchange for 10 and two old exchanges (10 and 15) still outstanding as their maturity is longer than one day, for a total outstanding amount of 35 on day 2. In day 3 instead there is a new exchange for 5 and an old exchange from day 1 of 10. That is, 15 of outstanding amount.
Here's a more visual explanation:
Monday Tuesday Wednesday
10 10 10 (Day 1, Value 10, matures in 3 days)
15 15 (Day 1, 15, 2 days)
10 (Day 2, 10, 1 day)
5 (Day 3, 5, 3 days with remainder not shown)
-------------------------------------
25 35 15 (Outstanding amount on each day)
Is there a simple way to get this result?
First of all in the main subquery we find SUM of all Values for current date. Then add to them values from previous dates according their MATURITY (the second subquery).
SQLFiddle demo
select T1.Date,T1.SumValue+
IFNULL((select SUM(VALUE)
from T
where
T1.Date between
T.Date+1 and T.Date+Maturity-1 )
,0)
FROM
(
select Date,
sum(Value) as SumValue
from T
group by Date
) T1
order by DATE
I'm not sure if this is what you are looking for, perhaps if you give more detail
select
DATE
,sum(VALUE) as OUTSTANDING_AMOUNT
from
NameOfYourTable
group by
DATE
Order by
DATE
I hope this helps
Each date considers each row for inclusion in the summation of value
SELECT d.DATE, SUM(m.VALUE) AS OUTSTANDING_AMOUNT
FROM yourTable AS d JOIN yourtable AS m ON d.DATE >= m.MATURITY
GROUP BY d.DATE
ORDER BY d.DATE
A possible solution with a tally (numbers) table
SELECT date, SUM(value) outstanding_amount
FROM
(
SELECT date + maturity - n.n date, value, maturity
FROM table1 t JOIN
(
SELECT 1 n UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5
) n ON n.n <= maturity
) q
GROUP BY date
Output:
| DATE | OUTSTANDING_AMOUNT |
-----------------------------
| 1 | 25 |
| 2 | 35 |
| 3 | 15 |
Here is SQLFiddle demo