I have a reports table with the following structure :
I want a SQL Query to get the report for the last 365 days by following conditions :
Group dates if the same date is repeated.
The days which the report is not available for the last 365 days, I need those days added to the result rows with 0 as their success and failed recipients.
I tried to get it by group by report dates
SELECT report_date, SUM(success_recipient) as success_recipient, SUM(failed_recipient) as failed_recipient FROM reports GROUP BY report_date;
and I have got the grouped result which satisfies the first condition
Now I need to append the rest of the days in the last 365 days to this result in which 0 as their success and failure recipients.
Expected result :
and so on ..
MYSQL VERSION : 5.6
One way to achieve this is using "with recursive" to generate all dates you need in you output and then outer join to the rest of your query. Note: I use the number 356 as it is in your description but it seems more appropriate to use date difference as this approach does not take into account leap years. Using the query below you will get NULL values in case you have no data. If you need the value 0 you can use coalesce(sum(...), 0).
with recursive
dates as (
select curdate()-356 dt
union all
select dt+1 from dates
where dt < curdate()
)
select
dt report_date,
sum(success_recipient) success_recipient,
sum(failed_recipient) failed_recipient
from dates
left join reports on report_date = dt
group by report_date;
From the above comments and the answer, I could write this query which gave me the expected outcome :
SELECT a.date, SUM(COALESCE(r.success_recipient, 0)), SUM(COALESCE(r.failed_recipient, 0))
FROM (
SELECT curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a) + (1000 * d.a) ) DAY AS date
FROM (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as d
) a
LEFT JOIN reports r ON a.date = r.report_date
WHERE a.date between DATE_SUB(CURDATE(), INTERVAL 1 YEAR) and now()
GROUP BY a.date;
Related
MySql version: 5.6.47
Assuming I have a table like this:
[from:Datetime][to:Datetime][amount:Decimal(10,2)]
[2020/01/15 ][2020/02/15 ][300 ]
I want to create a view like this out of it:
[period:char(7)][amount:Decimal(10,2)]
[2020/01 ][150 ]
[2020/02 ][150 ]
The from and to dates are split up in the singular months. The amount is multiplied with the amount of days in that particular month over the total amount of days between from and to. From and to could span n amount of month.
Is that even possible or am I wasting my time researching this?
Assuming that the amount of months in a range is not over 100:
SELECT id,
datefrom,
datetill,
amount,
monthstart,
monthfinish,
amount * (DATEDIFF(LEAST(datetill, monthfinish), GREATEST(datefrom, monthstart)) + 1) / (DATEDIFF(datetill, datefrom) + 1) monthamount
FROM ( SELECT test.*,
(test.datefrom - INTERVAL DAY(test.datefrom) - 1 DAY) + INTERVAL numbers.num MONTH monthstart,
LAST_DAY((test.datefrom - INTERVAL DAY(test.datefrom) - 1 DAY) + INTERVAL numbers.num MONTH) monthfinish
FROM test
JOIN ( SELECT t1.num*10+t2.num num
FROM (SELECT 0 num UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t1
JOIN (SELECT 0 num UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) t2
) numbers
HAVING monthstart <= test.datetill
AND monthfinish >= test.datefrom
) subquery
ORDER BY id, monthstart;
fiddle
PS. Don't be surprised if the total sum doesn't match in the last digit.
I need zero values also in group by clause .Have read almost all question related to this on Stackoverflow, but none of the solutions have worked.
My Table is
Need to get sum of score grouped by day of month.But I am not getting zero against the days not present in the table
SELECT SUM(engagement_score), DAY(creation_dt)
FROM qee_emp_engagement_index
RIGHT JOIN (
SELECT 1 AS index1 UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6 UNION ALL
SELECT 7 UNION ALL
SELECT 8 UNION ALL
SELECT 9) AS a ON a.index1 = DAY(creation_dt)
WHERE org_id = 1
GROUP BY a.index1
ORDER BY a.index1 ASC
I would write the query using a left join rather than a right join (the logic of left join makes more sense to me: keep all the rows in the first table). But, your problem is the where clause. That logic should go in the on clause:
SELECT COALESCE(SUM(eei.engagement_score), 0), a.index1
FROM (SELECT 1 AS index1 UNION ALL
SELECT 2 UNION ALL
SELECT 3 UNION ALL
SELECT 4 UNION ALL
SELECT 5 UNION ALL
SELECT 6 UNION ALL
SELECT 7 UNION ALL
SELECT 8 UNION ALL
SELECT 9
) a LEFT JOIN
qee_emp_engagement_index eei
ON a.index1 = DAY(eei.creation_dt) AND eei.org_id = 1
GROUP BY a.index1
ORDER BY a.index1 ASC;
In addition, the GROUP BY and SELECT should use the column from the driving table.
You can do right join between your table and a temp table containing all dates in a range.
Dates can be generated using below query
select a.Date
from (
select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date
from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
) a
where a.Date between '2015-01-01' and '2017-01-01'
SELECT DATE(creation_date), SUM(engagement_score) FROM table_name GROUP BY DATE(creation_date)
There is no need to do that weird right join in your code
If you want to group it by the day of month and not by the date, use DAY() instead of DATE()
EDIT:
If you want a zero when the day of month doesn't exist, try this:
SELECT d, SUM(score) FROM
((SELECT creation_date AS d, engagement_score AS score FROM table_name)
UNION ALL
(SELECT d, 0 AS engagement_score FROM table_name WHERE d BETWEEN 1 AND 31))
GROUP BY d
Table Pricing
What i want, is to provide a start_date & an end_date in order to get the price_per_day for each day between those 2 dates.
For example if i set as a start_date = 2015-05-30 & end_date = 2015-06-02, the desired output is
2015-05-30 | 24.00
2015-05-31 | 24.00
2015-06-01 | 27.00
2015-06-02 | 27.00
UPDATE:
Even if this output would be ok for me
24.00
24.00
27.00
27.00
You have to select dates greater than say 'From' date and lesser than 'To' date. I have posted the following without testing so please test and let me know in case of any errors.
SELECT Price FROM Pricing WHERE start_date >= '2015-05-30' AND end_date <= '2015-06-02'
Edit:
Please make sure the start_date and end_date have the same type as the dates provided. Just to be on the safe side, you could also convert them into datetime or convert(varchar, yourDatevariable, 103) but you have to apply it on both sides of the condition.
What you need to do is have a range of dates (ie, 1 row per date). This can be done a few ways. Probably the most efficient is having a calendar table, but if you cannot add new tables that is not possible.
You can have a table of numbers and add that to the starting date of each row, but again this requires a new table.
As such the option is to have a set of unioned queries to generate a range of numbers. For example the following will return 10 rows with the numbers 0 to 9:-
SELECT 0 a UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9
You can cross join such sub queries against each other to generate a larger range of numbers, and add that to your start date where the results is less than of equal to the end date:-
SELECT id, start_date, end_date, price_per_day, DATE_ADD(start_date, INTERVAL (units.a + tens.a * 10 + hundreds.a * 100) DAY) AS aDay
FROM pricing
CROSS JOIN (SELECT 0 a UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) units
CROSS JOIN (SELECT 0 a UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) tens
CROSS JOIN (SELECT 0 a UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) hundreds
WHERE DATE_ADD(start_date, INTERVAL (units.a + tens.a * 10 + hundreds.a * 100) DAY) <= end_date
The above will cope with up to 1000 days between the dates. Easy to expand to cope with 10000 days or more, but will become slower.
You can then just use that as a sub query when checking the date range you are interested in:-
SELECT aDay, price_per_day
FROM
(
SELECT id, start_date, end_date, price_per_day, DATE_ADD(start_date, INTERVAL (units.a + tens.a * 10 + hundreds.a * 100) DAY) AS aDay
FROM pricing
CROSS JOIN (SELECT 0 a UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) units
CROSS JOIN (SELECT 0 a UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) tens
CROSS JOIN (SELECT 0 a UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7 UNION SELECT 8 UNION SELECT 9) hundreds
WHERE DATE_ADD(start_date, INTERVAL (units.a + tens.a * 10 + hundreds.a * 100) DAY) <= end_date
) sub0
WHERE aDay BETWEEN '2015-05-30' AND '2015-06-02'
ORDER BY aDay
Your request should look something like this :
SELECT Price FROM Pricing WHERE start_date = 2015-05-30 AND end_date = 2015-06-02
But to print out everydate date between those two, i've no idea.
This might help you thought :
How to list all dates between two dates
Although I have researched similar other questions, however could not reach the solution by following those, hence posting my following question, and apologies for a long question in an attempt to make my question more clear.
The image shows my table structure.
I want to run such a query to extract the 3 information,i.e
userId, count(), Date(viewTime)
i.e the no of counts of id that a user has viewed on daily basis in a interval of last 14 days,
also show count as 0 if there are no records for a user on a particular day
select userId, count(userId), Date(viewTime) from user_views
where DATE(viewTime) between DATE_SUB(DATE(NOW()), INTERVAL 90 DAY) AND now()
group by userId, date(viewTime);
By using the above query I am getting only the non-zero records, see in the following image:
However I want to show count as 0 for those days when there are no transaction of users. How do I achieve this?
You need to generate the dates dynamically for this and then use left join. Also note that since you are displaying the user_id it might be needed a cross join of distinct user_id with the dynamically generated dates.
From my previous answers related to showing missing dates MySql Single Table, Select last 7 days and include empty rows
Here is one for your case
select
t1.user_id,
coalesce(t2.cnt,0) as cnt,
t1.view_date
from
(
select DATE_FORMAT(a.Date,'%Y-%m-%d') as view_date,
x.user_id
from (
select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as Date
from (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as a
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as b
cross join (select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all select 5 union all select 6 union all select 7 union all select 8 union all select 9) as c
) a,(select distinct user_id from user_views)x
where a.Date between DATE_SUB(DATE(NOW()), INTERVAL 90 DAY) AND now()
)t1
left join
(
select user_id, count(user_id) as cnt, Date(view_time) as view_time from user_views
where DATE(view_time) between DATE_SUB(DATE(NOW()), INTERVAL 90 DAY) AND now()
group by user_id, date(view_time)
)t2
on t2.view_time = t1.view_date
and t1.user_id = t2.user_id
order by t1.view_date,t1.user_id
http://sqlfiddle.com/#!2/4136e/5
This question already has answers here:
generate days from date range
(30 answers)
Closed 5 years ago.
What is the best way in mysql to generate a series of dates in a given range?
The application I have in mind is to write a report query that returns a row for every date, regardless of whether there is any data to report. In its simplest form:
select dates.date, sum(sales.amount)
from <series of dates between X and Y> dates
left join sales on date(sales.created) = dates.date
group by 1
I have tried creating a table with lots of dates, but that seems like a poor workaround.
if you're in a situation like me where creating temporary tables is prohibited, and setting variables is also not allowed, but you want to generate a list of dates in a specific period, say current year to do some aggregation, use this
select * from
(select adddate('1970-01-01',t4*10000 + t3*1000 + t2*100 + t1*10 + t0) gen_date from
(select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
(select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
(select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
(select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
(select 0 t4 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
where gen_date between '2017-01-01' and '2017-12-31'
I think having a calendar table is a good idea; you can gain a lot of reporting and query functionality, especially when filling sparse data ranges.
I found this article with what seems to be a good example.
You may use a variable generate date series:
Set #i:=0;
SELECT DATE(DATE_ADD(X,
INTERVAL #i:=#i+1 DAY) ) AS datesSeries
FROM yourtable, (SELECT #i:=0) r
where #i < DATEDIFF(now(), date Y)
;
Not sure if this is what you have tried :) though.
Next use above generated query as a table to left join:
set #i:=0;
select
d.dates,
sum(s.amount) as TotalAmount
from(
SELECT DATE(DATE_ADD(X,
INTERVAL #i:=#i+1 DAY) ) AS dateSeries
FROM Sales, (SELECT #i:=0) r
where #i < DATEDIFF(now(), date Y)
) dates d
left join Sales s
on Date(s.Created) = Date(d.dateSeries)
group by 1
;
You can also use Temporary Table to generate date series. Check below query:
CREATE TEMPORARY TABLE daterange (dte DATE);
SET #counter := -1;
WHILE (#counter < DATEDIFF(DATE(_todate), DATE(_fromdate))) DO
INSERT daterange VALUES (DATE_ADD(_fromdate, INTERVAL #counter:=#counter + 1 DAY));
END WHILE;
SELECT dates.dte, SUM(sales.amount)
FROM daterange dates
LEFT JOIN sales ON DATE(sales.created) = dates.date
GROUP BY dates.dte;