specific status on consecutive days - mysql

I have a MySQL table ATT which has EMP_ID,ATT_DATE,ATT_STATUS with ATT_STATUS with different values 1-Present,2-Absent,3-Weekly-off. I want to find out those EMP_ID's which have status 2 consecutively for 10 days in a given date range.
Please help


Please have a try with this:
SELECT EMP_ID FROM (
SELECT
IF((#prevDate!=(q.ATT_DATE - INTERVAL 1 DAY)) OR (#prevEmp!=q.EMP_ID) OR (q.ATT_STATUS != 2), #rownum:=#rownum+1, #rownum:=#rownum) AS rownumber, #prevDate:=q.ATT_DATE, #prevEmp:=q.EMP_ID, q.*
FROM (
SELECT
EMP_ID
, ATT_DATE
, ATT_STATUS
FROM
org_tb_dailyattendance, (SELECT #rownum:=0, #prevDate:='', #prevEmp:=0) vars
WHERE ATT_DATE BETWEEN '2013-01-01' AND '2013-02-15'
ORDER BY EMP_ID, ATT_DATE, ATT_STATUS
) q
) sq
GROUP BY EMP_ID, rownumber
HAVING COUNT(*) >= 10
The logic is, to first sort the table by employee id and the dates. Then introduce a rownumber which increases only if
the days are not consecutive or
the employee id is not the previous one or
the status is not 2
Then I just grouped by this rownumber and counted if there are 10 rows in each group. That should be the ones who were absent for 10 days or more.


Have you tried something like this
SELECT EMP_ID count(*) as consecutive_count min(ATT_DATE)
FROM (SELECT * FROM ATT ORDER BY EMP_ID)
GROUP BY EMP_ID, ATT_DATE
WHERE ATT_STATUS = 2
HAVING consecutive_count > 10

Related

Get sum of previous records in query and add or subtract the following results

Case:
I select an initial date and an end date, it should bring me the movements of all the products in that date range, but if there were movements before the initial date (records in table), I want to obtain the previous sum (prevData)
if the first move is exit 5 and the second move is income 2.
I would have in the first row (prevData-5), second row would have (prevData-5 + 2) and thus have a cumulative.
The prevData would be calculated as the sum of the above, validating the product id of the record, I made the query but if the product has 10 movements, I would do the query 10 times, and how would I identify the sum of another product_id?
SELECT
ik.id,
ik.quantity,
ik.date,
ik.product_id,
#balance = (SELECT SUM(quantity) FROM table_kardex WHERE product_id = ik.product_id AND id < ik.id)
from table_kardex ik
where ik.date between '2021-11-01' and '2021-11-15'
order by ik.product_id,ik.id asc
I hope you have given me to understand, I will be attentive to any questions.
#table_kardex
id|quantity|date|product_id
1 8 2020-10-12 2
2 15 2020-10-12 1
3 5 2021-11-01 1
4 10 2021-11-01 2
5 -2 2021-11-02 1
6 -4 2021-11-02 2
#result
id|quantity|date|product_id|saldo
3 5 2021-11-01 1 20 (15+5)
5 -2 2021-11-02 1 18 (15+5-2)
4 10 2021-11-01 2 18 (8+10-4)
6 -4 2021-11-02 2 14 (15+5-2)
Use MySQL 5.7

If you're using MySQL 8+, then analytic functions can be used here:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY product_id ORDER BY date) rn,
SUM(quantity) OVER (PARTITION BY product_id ORDER BY date) saldo
FROM table_kardex
WHERE date BETWEEN '2021-11-01' AND '2021-11-15'
)
SELECT id, quantity, date, product_id, saldo
FROM cte
WHERE rn > 1
ORDER BY product_id, date;

MySQL 5.7
Try this:
SELECT *
FROM (
SELECT product_id,
t1.`date`,
SUM(t2.quantity) - t1.quantity cumulative_quantity_before,
SUM(t2.quantity) cumulative_quantity_after
FROM table t1
JOIN table t2 USING (product_id)
WHERE t1.`date` >= t2.`date`
AND t1.`date` <= #period_end
GROUP BY product_id, t1.`date`, t1.quantity
) prepare_data
WHERE `date` >= #period_start;

The easiest solution is to use the window function SUM OVER to get the running total. In the second step reduce this to the date you want to have this started:
SELECT id, quantity, date, product_id, balance
FROM
(
SELECT
id,
quantity,
date,
product_id,
SUM(quantity) OVER (PARTITION BY product_id ORDER BY id) AS balance
from table_kardex ik
where date < DATE '2021-11-16'
) cumulated
WHERE date >= DATE '2021-11-01'
ORDER BY product_id, id;
UPDATE: You have changed your request to mention that you are using an old MySQL version (5.7). This doesn't support window functions. In that case use your original query. If I am not mistaken, though, #balance = (...) is invalid syntax for MySQL. And according to your explanation you want id <= ik.id, not id < ik.id:
SELECT
ik.id,
ik.quantity,
ik.date,
ik.product_id,
(
SELECT SUM(quantity)
FROM table_kardex
WHERE product_id = ik.product_id AND id <= ik.id
) AS balance
FROM table_kardex ik
WHERE ik.date >= DATE '2021-11-01' AND ik.date < DATE '2021-11-16'
ORDER BY ik.product_id, ik.id;
The appropriate indexes for this query are:
create index idx1 on table_kardex (date, product_id, id);
create index idx2 on table_kardex (product_id, id, quantity);

SELECT the first date in which 3 consecutive entries are between two dates

I would like to know the first date of the first 3 consecutive entries that are between two dates. Based on my SQLFiddle, I would expect the output to be '2021-01-24'.
I've looked at many examples but can't get them to work.
This query is not working how I want it to, I can't figure out the missing piece of my query. Here is the SQLFIDDLE: http://sqlfiddle.com/#!9/935fbd/1
SELECT DISTINCT
logDate
FROM
FoodLog
WHERE
studentID = '1329' AND logDate BETWEEN '2021-01-01' AND '2021-05-01'
GROUP BY
logDate
HAVING
COUNT(logDate) = 3
I've tried working with the following, but can't figure out how to limit the search to studentID='1329' or my date range:
SELECT DISTINCT
f.id,
f.logDate
FROM
FoodLog f,
(
SELECT
f1.logDate START,
f2.logDate NEXT
FROM
FoodLog f1,
FoodLog f2
WHERE
f2.logDate <= DATE_ADD(f1.logDate, INTERVAL 1 DAY) AND f2.logDate > f1.logDate
) f2
WHERE
f.logDate = f2.start OR(
f.logDate = f2.next AND f2.start IS NOT NULL
)
LIMIT 1

WITH
cte1 AS (
SELECT DISTINCT logDate
FROM FoodLog
WHERE logDate BETWEEN '2021-01-01' AND '2021-05-01'),
cte2 AS (
SELECT logDate, LEAD(logDate, 2) OVER (ORDER BY logDate) next2date
FROM cte1
)
SELECT MIN(logDate) logDate
FROM cte2
WHERE DATEDIFF(next2date, logDate) = 2;
fiddle

Calculating a Moving Average MySQL?

Good Day,
I am using the following code to calculate the 9 Day Moving average.
SELECT SUM(close)
FROM tbl
WHERE date <= '2002-07-05'
AND name_id = 2
ORDER BY date DESC
LIMIT 9
But it does not work because it first calculates all of the returned fields before the limit is called. In other words it will calculate all the closes before or equal to that date, and not just the last 9.
So I need to calculate the SUM from the returned select, rather than calculate it straight.
IE. Select the SUM from the SELECT...
Now how would I go about doing this and is it very costly or is there a better way?

If you want the moving average for each date, then try this:
SELECT date, SUM(close),
(select avg(close) from tbl t2 where t2.name_id = t.name_id and datediff(t2.date, t.date) <= 9
) as mvgAvg
FROM tbl t
WHERE date <= '2002-07-05' and
name_id = 2
GROUP BY date
ORDER BY date DESC
It uses a correlated subquery to calculate the average of 9 values.

Starting from MySQL 8, you should use window functions for this. Using the window RANGE clause, you can create a logical window over an interval, which is very powerful. Something like this:
SELECT
date,
close,
AVG (close) OVER (ORDER BY date DESC RANGE INTERVAL 9 DAY PRECEDING)
FROM tbl
WHERE date <= DATE '2002-07-05'
AND name_id = 2
ORDER BY date DESC
For example:
WITH t (date, `close`) AS (
SELECT DATE '2020-01-01', 50 UNION ALL
SELECT DATE '2020-01-03', 54 UNION ALL
SELECT DATE '2020-01-05', 51 UNION ALL
SELECT DATE '2020-01-12', 49 UNION ALL
SELECT DATE '2020-01-13', 59 UNION ALL
SELECT DATE '2020-01-15', 30 UNION ALL
SELECT DATE '2020-01-17', 35 UNION ALL
SELECT DATE '2020-01-18', 39 UNION ALL
SELECT DATE '2020-01-19', 47 UNION ALL
SELECT DATE '2020-01-26', 50
)
SELECT
date,
`close`,
COUNT(*) OVER w AS c,
SUM(`close`) OVER w AS s,
AVG(`close`) OVER w AS a
FROM t
WINDOW w AS (ORDER BY date DESC RANGE INTERVAL 9 DAY PRECEDING)
ORDER BY date DESC
Leading to:
date |close|c|s |a |
----------|-----|-|---|-------|
2020-01-26| 50|1| 50|50.0000|
2020-01-19| 47|2| 97|48.5000|
2020-01-18| 39|3|136|45.3333|
2020-01-17| 35|4|171|42.7500|
2020-01-15| 30|4|151|37.7500|
2020-01-13| 59|5|210|42.0000|
2020-01-12| 49|6|259|43.1667|
2020-01-05| 51|3|159|53.0000|
2020-01-03| 54|3|154|51.3333|
2020-01-01| 50|3|155|51.6667|

Use something like
SELECT
sum(close) as sum,
avg(close) as average
FROM (
SELECT
(close)
FROM
tbl
WHERE
date <= '2002-07-05'
AND name_id = 2
ORDER BY
date DESC
LIMIT 9 ) temp
The inner query returns all filtered rows in desc order, and then you avg, sum up those rows returned.
The reason why the query given by you doesn't work is due to the fact that the sum is calculated first and the LIMIT clause is applied after the sum has already been calculated, giving you the sum of all the rows present

an other technique is to do a table:
CREATE TABLE `tinyint_asc` (
`value` tinyint(3) unsigned NOT NULL default '0',
PRIMARY KEY (value)
) ;
​
INSERT INTO `tinyint_asc` 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),(24),(25),(26),(27),(28),(29),(30),(31),(32),(33),(34),(35),(36),(37),(38),(39),(40),(41),(42),(43),(44),(45),(46),(47),(48),(49),(50),(51),(52),(53),(54),(55),(56),(57),(58),(59),(60),(61),(62),(63),(64),(65),(66),(67),(68),(69),(70),(71),(72),(73),(74),(75),(76),(77),(78),(79),(80),(81),(82),(83),(84),(85),(86),(87),(88),(89),(90),(91),(92),(93),(94),(95),(96),(97),(98),(99),(100),(101),(102),(103),(104),(105),(106),(107),(108),(109),(110),(111),(112),(113),(114),(115),(116),(117),(118),(119),(120),(121),(122),(123),(124),(125),(126),(127),(128),(129),(130),(131),(132),(133),(134),(135),(136),(137),(138),(139),(140),(141),(142),(143),(144),(145),(146),(147),(148),(149),(150),(151),(152),(153),(154),(155),(156),(157),(158),(159),(160),(161),(162),(163),(164),(165),(166),(167),(168),(169),(170),(171),(172),(173),(174),(175),(176),(177),(178),(179),(180),(181),(182),(183),(184),(185),(186),(187),(188),(189),(190),(191),(192),(193),(194),(195),(196),(197),(198),(199),(200),(201),(202),(203),(204),(205),(206),(207),(208),(209),(210),(211),(212),(213),(214),(215),(216),(217),(218),(219),(220),(221),(222),(223),(224),(225),(226),(227),(228),(229),(230),(231),(232),(233),(234),(235),(236),(237),(238),(239),(240),(241),(242),(243),(244),(245),(246),(247),(248),(249),(250),(251),(252),(253),(254),(255);
After you can used it like that:
select
date_add(tbl.date, interval tinyint_asc.value day) as mydate,
count(*),
sum(myvalue)
from tbl inner
join tinyint_asc.value <= 30 -- for a 30 day moving average
where date( date_add(o.created_at, interval tinyint_asc.value day ) ) between '2016-01-01' and current_date()
group by mydate

This query is fast:
select date, name_id,
case #i when name_id then #i:=name_id else (#i:=name_id)
and (#n:=0)
and (#a0:=0) and (#a1:=0) and (#a2:=0) and (#a3:=0) and (#a4:=0) and (#a5:=0) and (#a6:=0) and (#a7:=0) and (#a8:=0)
end as a,
case #n when 9 then #n:=9 else #n:=#n+1 end as n,
#a0:=#a1,#a1:=#a2,#a2:=#a3,#a3:=#a4,#a4:=#a5,#a5:=#a6,#a6:=#a7,#a7:=#a8,#a8:=close,
(#a0+#a1+#a2+#a3+#a4+#a5+#a6+#a7+#a8)/#n as av
from tbl,
(select #i:=0, #n:=0,
#a0:=0, #a1:=0, #a2:=0, #a3:=0, #a4:=0, #a5:=0, #a6:=0, #a7:=0, #a8:=0) a
where name_id=2
order by name_id, date
If you need an average over 50 or 100 values, it's tedious to write, but
worth the effort. The speed is close to the ordered select.

least value in count

i have a table employee(id,dept_id,salary,hire_date,job_id) . the following query i have to execute.
Show all the employee who were hired on the day of the week on which least no of employee were hired.
i have done the query, but am not able to get the least. please check if am correct.
select id, WEEKDAY(hire_date)+1 as days,count(WEEKDAY(hire_date)+1) as count
from test.employee group by days

This should get you the weekday on which the least number of employees were hired:
SELECT
count(id) as `Total`,
WEEKDAY(hire_date) as `DoW`
FROM
test.employee
GROUP BY `DoW`
ORDER BY `Total` DESC LIMIT 1;

select id from test.employee where hire_date in
( select count(id) count,hire_date
from test.employee
order by count desc
limit 1)
this should work

You may try this, as it will not limit to one record if you have multiple week days where the same least number of employees were hired. In reality it makes sense. The following is based on sample data.
Query:
-- find minimum id count
SELECT MIN(e.counts) INTO #min
FROM (SELECT COUNT(*) as counts,
WEEKDAY(hire_date+1) as day
FROM employee
GROUP BY WEEKDAY(hire_date+1)) e
;
-- show weekdays with minimum id counts
SELECT e2.counts as mincount,
WEEKDAY(e1.hire_date+1) as weekday
FROM employee e1
JOIN (SELECT COUNT(id) as counts,
WEEKDAY(hire_date+1) as day
FROM employee
GROUP BY day
HAVING COUNT(*) = #min) e2
ON WEEKDAY(e1.hire_date+1) = e2.day;
Results:
MINCOUNT WEEKDAY
1 6
1 3
1 4
1 2
SQLFIDDLE

select min(id), WEEKDAY(hire_date)+1 as days,count(WEEKDAY(hire_date)+1) as count
from test.employee group by days

SELECT
*
FROM
employee
WHERE
DAYOFWEEK(hire_date)
IN
(
SELECT
weekday
FROM
(
SELECT
count(*) as bcount,
DAYOFWEEK(hire_date) as weekday
FROM
employee as a
GROUP BY
weekday
HAVING
bcount = (
SELECT
MIN(tcount)
FROM
(
SELECT
count(*) as tcount,
DAYOFWEEK(hire_date) as weekday
FROM
employee
GROUP BY
weekday
) as t
)
) as q

MySQL group by where day <= x

I need some help figuring out the correct SQL statement.
If've got a table with the following structure:
id, product_id, units, timestamp
I wan't a list which contains the all over units per day. A product has maximum one record per day.
So my first try was:
SELECT
DATE(timestamp) as day, SUM(units) as overall_units
FROM
tbl
GROUP BY
DATE(timestamp);
Normally this should do it. But sometimes there are days where is no record for a product. Nevertheless the units are still in the warehouse so they should be in the calculation.
For example:
We have 3 products. Cars, pens and wheels.
Records from 2012-10-20:
Cars => 5
Pens => 20
Wheels => 4
Records from 2012-10-21
Cars => 5
Wheels => 6
My query would give the following results:
2012-10-20 => 29
2012-10-21 => 11
But I want, that if there's no record for a product for the day it should use the record for this product which is the nearest one back in time.
So it should be:
2012-10-21 => 31
I hope you understand my needs.

SELECT
MAX (DATE(timestamp) ) as day, SUM(units) as overall_units
FROM tbl
update ::
SELECT max(day),sum(ou) from
( select DATE(timestamp) as day, SUM(units) as ou
FROM tbl
GROUP BY DATE(timestamp);
)
inner qry will return
2012-10-20 , 29
2012-10-21 , 11
and the final query will return
2012-10-21 , 40

SELECT
dd.ddate AS day
, SUM(t.units) as overall_units
FROM
( SELECT DISTINCT
product_id
FROM
tbl
) AS dp
CROSS JOIN
( SELECT DISTINCT
DATE(timestamp) AS ddate
FROM
tbl
) AS dd
JOIN
tbl AS t
ON t.id =
( SELECT
tt.id
FROM
tbl AS tt
WHERE
tt.product_id = dp.product_id
AND
tt.timestamp < dd.ddate + INTERVAL 1 DAY
ORDER BY tt.timestamp DESC
LIMIT 1
)
GROUP BY
dd.ddate ;

I think you should look into the DISTINCT() function.
The query could be something like: SELECT DISTINCT(product_id), * FROM tbl ORDER BY timestamp DESC; The use PHP to loop through your results and cumulate the units.