I have the two following (MySQL) tables called "Jobs" and "Employees_Jobs:
Jobs:
job_id job_creation_date
1 2016-01-01
2 2016-01-02
Employees_Jobs (job applications):
EJ_job_id EJ_creation_date
1 2016-01-02
2 2016-01-02
2 2016-01-03
I want MySQL returning the number of jobs created, and the number of job applications created per day of the week; taking the above data it should return:
weekday num_of_jobs_entered num_of_applications_entered
Friday 1 0
Saturday 1 2 // corrected from 1
Sunday 0 1 // 2
I now have the following query:
SELECT
DAYNAME(job_creation_date) as weekday,
(SELECT COUNT(*) FROM Jobs) as num_of_jobs_entered,
(SELECT COUNT(*) FROM Employees_Jobs) as num_of_applications_entered
FROM
dual
GROUP BY
weekday
ORDER BY
weekday;
What am I doing wrong?
Thanks!
Try this:
SELECT wd, COUNT(jcnt), COUNT(ecnt) FROM
(SELECT DAYNAME(job_creation_date) wd, 1 jcnt, null ecnt FROM Jobs UNION ALL
SELECT DAYNAME(EJ_creation_date), null, 1 FROM Employees_Jobs ) a
GROUP BY wd
See here for a working example.
EDIT
If I understand you correctly you want to divide the above calculated counts of job offers and requirements by the current week number of the current year (which will make sense of course only if all considered job counts have also occurred in the current year -> some filtering might become necessary).
However, whithout the filtering you could do
SELECT week_day, Week(CURDATE()) weekCurrdate,
COUNT(num_of_jobs_entered)/Week(CURDATE()) avg_of_jobs_entered,
COUNT(num_of_applications_entered)/Week(CURDATE()) avg_num_of_applications_entered
FROM (
SELECT DAYNAME(job_creation_date) week_day, 1 num_of_jobs_entered,
null num_of_applications_entered FROM Jobs UNION ALL
SELECT DAYNAME(EJ_creation_date), null, 1 FROM Employees_Jobs ) A
GROUP BY week_day;
The IFNULL() function is obsolete but you will have to use COUNT() instead in the outer select. Since the current week number is 1 at the moment the result from this query will (presently!) be identical to the one of the previous query, see modified fiddle here.
Related
i have an orders table, and i need to fetch the orders record by month. but i have terms if there is no data in a month it should still show the data but forcing to zero like this:
what i have done is using my query:
select sum(total) as total_orders, DATE_FORMAT(created_at, "%M") as date
from orders
where is_active = 1
AND tenant_id = 2
AND created_at like '%2021%'
group by DATE_FORMAT(created_at, "%m")
but the result is only fetched the existed data:
can anyone here help me to create the exactly query?
Thank you so much
Whenever you're trying to use a value that doesn't exist in the table, one option is to use a reference; whether it's from a table or a query-generated value.
I'm guessing that in terms of date data, the column created_at in table orders may have a complete list all the 12 months in a year regardless of which year.
Let's assume that the table data for orders spans from 2019 to present date. With that you can simply create a 12 months reference table for a LEFT JOIN operation. So:
SELECT MONTHNAME(created_at) mnt FROM orders GROUP BY MONTHNAME(created_at);
You can append that into your query like:
SELECT IFNULL(SUM(total),0) as total_orders, mnt
from (SELECT MONTHNAME(created_at) mnt FROM orders GROUP BY MONTHNAME(created_at)) mn
LEFT JOIN orders o
ON mn.mnt=MONTHNAME(created_at)
AND is_active = 1
AND tenant_id = 2
AND created_at like '%2021%'
GROUP BY mnt;
Apart from adding the 12 months sub-query and a LEFT JOIN, there are 3 other changes from your original query:
IFNULL() is added to the SUM() operation in SELECT to return 0 if the value is non-existent.
All the WHERE conditions has been switched to ON since remaining it as WHERE will make the LEFT JOIN becoming a normal JOIN.
GROUP BY is using the sub-query generated month (mnt) value instead.
Taking consideration of table orders might not have the full 12 months, you can generate it from query. There are a lot of ways of doing it but here I'm only going to show the UNION method that works with most MySQL version.
SELECT MONTHNAME(CONCAT_WS('-',YEAR(NOW()),mnt,'01')) dt
FROM
(SELECT 1 AS mnt UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION
SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION
SELECT 9 UNION SELECT 10 UNION SELECT 11 UNION SELECT 12) mn
If you're using MariaDB version that supports SEQUENCE ENGINE, the same query above is much shorter:
SELECT MONTHNAME(CONCAT_WS('-',YEAR(NOW()),mnt,'01'))
FROM (SELECT seq AS mnt FROM seq_1_to_12) mn
I'm using MariaDB 10.5 in this demo fiddle however it seems like the month name ordering is based on the name value rather than the month itself so it looks un-ordered. It's in the correct order if it's in MySQL 8.0 fiddle though.
Thanks all for the answers & comments i really appreciate it.
i solved it by create table helper for static months then use union and aliasing, since i need the months in indonesia, i create case-when function too.
so, the query is like this:
SELECT total_orders,
(CASE date WHEN 01 THEN 'Januari'
WHEN 02 THEN 'Februari'
WHEN 03 THEN 'Maret'
WHEN 04 THEN 'April'
WHEN 05 THEN 'Mei'
WHEN 06 THEN 'Juni'
WHEN 07 THEN 'Juli'
WHEN 08 THEN 'Agustus'
WHEN 09 THEN 'September'
WHEN 10 THEN 'Oktober'
WHEN 11 THEN 'November'
WHEN 12 THEN 'Desember'
ELSE date END ) AS date
FROM (SELECT SUM(total) AS total_orders,
DATE_FORMAT(created_at, "%m") AS date
FROM orders
WHERE is_active = 1
AND tenant_id = 2
AND created_at like '%2021%'
GROUP BY DATE_FORMAT(created_at, "%m")
UNION
SELECT 0 AS total_orders,
code AS date
FROM quantum_default_months ) as Q
GROUP BY date
I still don't know if this query is fully correct or not, but I get my exact result.
cmiiw.
thanks all
I need an Amazon Redshift SQL query to calculate the number of a particular day fall in between two dates.
Date Format - YYYY-MM-DD
For example - Start date = 2019-06-14, End Date = 2019-10-09, Day - 2nd of every month
Now, I want to calculate the count of 2nd-day fall in between 2019-06-14 and 2019-10-09
So, the actual result for the above example should be 4. Since 4 times the 2nd-day will fall in between 2019-06-14 and 2019-10-09.
I tried the DATE_DIFF function and months_between function of redshift. But failed to build the logic. Since not able to understand what math or equation should be.
for me it seems as if you wanted to select from a calendar table. That's how you can solve your problem. You'll notice that the query looks a little hacky because Redshift does not support any functions to generate sequences, which leaves you with creating sequence tables yourself (see seq_10 and seq_1000). Once you have a sequence, you can easily create a calendar with all the information you need (eg. day_of_month).
That's the query answering your question:
WITH seq_10 as (
SELECT 1 UNION ALL
SELECT 1 UNION ALL
SELECT 1 UNION ALL
SELECT 1 UNION ALL
SELECT 1 UNION ALL
SELECT 1 UNION ALL
SELECT 1 UNION ALL
SELECT 1 UNION ALL
SELECT 1 UNION ALL
SELECT 1
), seq_1000 as (
select
row_number() over () - 1 as n
from
seq_10 a cross join
seq_10 b cross join
seq_10 c
), calendar as (
select '2018-01-01'::date + n as date,
extract(day from date) as day_of_month,
extract(dow from date) as day_of_week
from seq_1000
)
select count(*) from calendar
where day_of_month = 2
and date between '2019-06-14' and '2019-10-09'
I have this table called rsvp which has 3 columns:
id
name
time
The time will either be 6:00pm, 6:30pm, 7:00pm, 7:30pm
What I am trying to do is get a query that will return 2 columns. Each of the times and the number of times these times are used.
For example, if I have two items in my table. One had a time of 6:00pm and the other is 7:00pm
My query would return:
6:00pm - 1
6:30pm - 0
7:00pm - 1
7:30pm - 0
How would I get these columns from a mysql query?
Here is one way to do this:
SELECT timelist.time, COUNT(rsvp.id) FROM (
SELECT '6:00pm' AS time UNION ALL
SELECT '6:30pm' UNION ALL
SELECT '7:00pm' UNION ALL
SELECT '7:30pm'
) AS timelist
LEFT JOIN rsvp ON timelist.time = rsvp.time
GROUP BY timelist.time
ORDER BY timelist.time
SELECT COUNT(id) as count, time FROM rsvp GROUP BY time
I have two tables
Students table :
id studentname admissionno
3 test3 3
2 test2 2
1 test 1
2nd table is fee :
id studentid created
1 3 2015-06-06 22:55:34
2 2 2015-05-07 13:32:48
3 1 2015-06-07 17:47:46
I need to fetch the students who haven't paid for the current month,
I'm performing the following query:
SELECT studentname FROM students
WHERE studentname != (select students.studentname from students
JOIN submit_fee
ON (students.id=submit_fee.studentid)
WHERE MONTH(CURDATE()) = MONTH(submit_fee.created)) ;
and I'm getting error:
'#1242 - Subquery returns more than 1 row'
Can you tell me what the correct query is to fetch all the students who haven't paid for the current month?
Use not in, please try query below :
SELECT s.*
FROM students s
WHERE s.id NOT IN ( SELECT sf.studentid FROM studentfees sf WHERE month(sf.created) = EXTRACT(month FROM (NOW())) )
You want to use not exists or a left join for this:
select s.*
from students s
where not exists (select 1
from studentfees sf
where s.id = sf.studentid and
sf.created >= date_sub(curdate(), interval day(curdate) - 1) and
sf.created < date_add(date_sub(curdate(), interval day(curdate) - 1), 1 month)
)
Note the careful construction of the date arithmetic. All the functions are on curdate() rather than on created. This allows MySQL to use an index for the where clause, if one is appropriate. One error in your query is the use of MONTH() without using YEAR(). In general, the two would normally be used together, unless you really want to combine months from different years.
Also, note that paying or not paying for the current month may not really answer the question. What if a student paid for the current month but missed the previous month's payment?
Data:
values date
14 1.1.2010
20 1.1.2010
10 2.1.2010
7 4.1.2010
...
sample query about january 2010 should get 31 rows. One for every day. And values vould be added. Right now I could do this with 31 queries but I would like this to work with one. Is it possible?
results:
1. 34
2. 10
3. 0
4. 7
...
This is actually surprisingly difficult to do in SQL. One way to do it is to have a long select statement with UNION ALLs to generate the numbers from 1 to 31. This demonstrates the principle but I stopped at 4 for clarity:
SELECT MonthDate.Date, COALESCE(SUM(`values`), 0) AS Total
FROM (
SELECT 1 AS Date UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
--
SELECT 28 UNION ALL
SELECT 29 UNION ALL
SELECT 30 UNION ALL
SELECT 31) AS MonthDate
LEFT JOIN Table1 AS T1
ON MonthDate.Date = DAY(T1.Date)
AND MONTH(T1.Date) = 1 AND YEAR(T1.Date) = 2010
WHERE MonthDate.Date <= DAY(LAST_DAY('2010-01-01'))
GROUP BY MonthDate.Date
It might be better to use a table to store these values and join with it instead.
Result:
1, 34
2, 10
3, 0
4, 7
Given that for some dates you have no data, you'll need to fill in the gaps. One approach to this is to have a calendar table prefilled with all dates you need, and join against that.
If you want the results to show day numbers as you have showing in your question, you could prepopulate these in your calendar too as labels.
You would join your data table date field to the date field of the calendar table, group by that field, and sum values. You might want to specify limits for the range of dates covered.
So you might have:
CREATE TABLE Calendar (
label varchar,
cal_date date,
primary key ( cal_date )
)
Query:
SELECT
c.label,
SUM( d.values )
FROM
Calendar c
JOIN
Data_table d
ON d.date_field = c.cal_date
WHERE
c.cal_date BETWEEN '2010-01-01' AND '2010-01-31'
GROUP BY
d.date_field
ORDER BY
d.date_field
Update:
I see you have datetimes rather than dates. You could just use the MySQL DATE() function in the join, but that would probably not be optimal. Another approach would be to have start and end times in the Calendar table defining a 'time bucket' for each day.
This works for me... Its a modification of a query I found on another site. The "INTERVAL 1 MONTH" clause ensures I get the current month data, including zeros for days that have no hits. Change this to "INTERVAL 2 MONTH" to get last months data, etc.
I have a table called "payload" with a column "timestamp" - Im then joining the timestamp column on to the dynamically generated dates, casting it so that the dates match in the ON clause.
SELECT `calendarday`,COUNT(P.`timestamp`) AS `cnt` FROM
(SELECT #tmpdate := DATE_ADD(#tmpdate, INTERVAL 1 DAY) `calendarday`
FROM (SELECT #tmpdate :=
LAST_DAY(DATE_SUB(CURDATE(),INTERVAL 1 MONTH)))
AS `dynamic`, `payload`) AS `calendar`
LEFT JOIN `payload` P ON DATE(P.`timestamp`) = `calendarday`
GROUP BY `calendarday`
To dynamically get the dates within a date range using SQL you can do this (example in mysql):
Create a table to hold the numbers 0 through 9.
CREATE TABLE ints ( i tinyint(4) );
insert into ints (i)
values (0),(1),(2),(3),(4),(5),(6),(7),(8),(9);
Run a query like so:
select ((curdate() - interval 2 year) + interval (t.i * 100 + u.i * 10 + v.i) day) AS Date
from
ints t
join ints u
join ints v
having Date between '2015-01-01' and '2015-05-01'
order by t.i, u.i, v.i
This will generate all dates between Jan 1, 2015 and May 1, 2015.
Output
2015-01-01
2015-01-02
2015-01-03
2015-01-04
2015-01-05
2015-01-06
...
2015-05-01
The query joins the table ints 3 times and gets an incrementing number (0 through 999). It then adds this number as a day interval starting from a certain date, in this case a date 2 years ago. Any date range from 2 years ago and 1,000 days ahead can be obtained with the example above.
To generate a query that generates dates for more than 1,000 days simply join the ints table once more to allow for up to 10,000 days of range, and so forth.
If I'm understanding the rather vague question correctly, you want to know the number of records for each date within a month. If that's true, here's how you can do it:
SELECT COUNT(value_column) FROM table WHERE date_column LIKE '2010-01-%' GROUP BY date_column