This is how my table looks like
-----------------------
posts
----------------------
id
created_at
..
..
How should the MySQL Query look like, so that i get the number of entries the last 7 days.
The result should look something like that:
['Mon' => 234, 'Tues' => 12, ...]
you can use datediff for that
select count(*), extract(day from created_at) current_day
from posts
where datediff(now(), created_at) <= 7
group by current_day;
If you want 0 values for other day, then you would have to use a left join on a fake table that generate last seven days.
select count(posts.created_at) as nb_occurence, c.a as number_of_day
from
(select b.a
from (select 1 a
union all select 2 a
union all select 3 a
union all select 4 a
union all select 5 a
union all select 6 a
union all select 7 a) b) c
left join posts on extract(day from posts.created_at) % 7 - c.a in (select extract(day from date_sub(created_at, INTERVAL 7 DAY)) % 7 from posts)
group by c.a;
SELECT COUNT(*)
FROM posts
WHERE created_at<xy
GROUP BY created_at
SELECT COUNT(*), DATE_FORMAT(created_at,'%a')
FROM posts
WHERE created_at <= NOW() AND created_at >= DATE_SUB(created_at, INTERVAL 7 DAY)
GROUP BY DATE_FORMAT(created_at,'%a')
To add 0 for weekday having null in count:
SELECT a.weekday, IFNULL(b.total,0) FROM
(SELECT 'Mon' as weekday from dual union SELECT 'Tue' as weekday from dual union SELECT 'Wed' as weekday from dual union SELECT 'Thu' as weekday from dual union SELECT 'Fri' as weekday from dual union SELECT 'Sat' as weekday from dual union SELECT 'Sun' as weekday from dual) a
LEFT JOIN
(SELECT COUNT(*) as total, DATE_FORMAT(created_at,'%a') as weekday
FROM posts
WHERE created_at <= NOW() AND created_at >= DATE_SUB(created_at, INTERVAL 7 DAY)
GROUP BY DATE_FORMAT(created_at,'%a')) b on a.weekday=b.weekday
Related
I have a table with sell orders and I want to list the COUNT of sell orders per day, between two dates, without leaving date gaps.
This is what I have currently:
SELECT COUNT(*) as Norders, DATE_FORMAT(date, "%M %e") as sdate
FROM ORDERS
WHERE date <= NOW()
AND date >= NOW() - INTERVAL 1 MONTH
GROUP BY DAY(date)
ORDER BY date ASC;
The result I'm getting is as follows:
6 May 1
14 May 4
1 May 5
8 Jun 2
5 Jun 15
But what I'd like to get is:
6 May 1
0 May 2
0 May 3
14 May 4
1 May 5
0 May 6
0 May 7
0 May 8
.....
0 Jun 1
8 Jun 2
.....
5 Jun 15
Is that possible?
Creating a range of dates on the fly and joining that against you orders table:-
SELECT sub1.sdate, COUNT(ORDERS.id) as Norders
FROM
(
SELECT DATE_FORMAT(DATE_SUB(NOW(), INTERVAL units.i + tens.i * 10 + hundreds.i * 100 DAY), "%M %e") as sdate
FROM (SELECT 0 i 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 i 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 i 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_SUB(NOW(), INTERVAL units.i + tens.i * 10 + hundreds.i * 100 DAY) BETWEEN DATE_SUB(NOW(), INTERVAL 1 MONTH) AND NOW()
) sub1
LEFT OUTER JOIN ORDERS
ON sub1.sdate = DATE_FORMAT(ORDERS.date, "%M %e")
GROUP BY sub1.sdate
This copes with date ranges of up to 1000 days.
Note that it could be made more efficient easily depending on the type of field you are using for your dates.
EDIT - as requested, to get the count of orders per month:-
SELECT aMonth, COUNT(ORDERS.id) as Norders
FROM
(
SELECT DATE_FORMAT(DATE_SUB(NOW(), INTERVAL months.i MONTH), "%Y%m") as sdate, DATE_FORMAT(DATE_SUB(NOW(), INTERVAL months.i MONTH), "%M") as aMonth
FROM (SELECT 0 i 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 UNION SELECT 10 UNION SELECT 11)months
WHERE DATE_SUB(NOW(), INTERVAL months.i MONTH) BETWEEN DATE_SUB(NOW(), INTERVAL 12 MONTH) AND NOW()
) sub1
LEFT OUTER JOIN ORDERS
ON sub1.sdate = DATE_FORMAT(ORDERS.date, "%Y%m")
GROUP BY aMonth
You are going to need to generate a virtual (or physical) table, containing every date in the range.
That can be done as follows, using a sequence table.
SELECT mintime + INTERVAL seq.seq DAY AS orderdate
FROM (
SELECT CURDATE() - INTERVAL 1 MONTH AS mintime,
CURDATE() AS maxtime
FROM obs
) AS minmax
JOIN seq_0_to_999999 AS seq ON seq.seq < TIMESTAMPDIFF(DAY,mintime,maxtime)
Then, you join this virtual table to your query, as follows.
SELECT IFNULL(orders.Norders,0) AS Norders, /* show zero instead of null*/
DATE_FORMAT(alldates.orderdate, "%M %e") as sdate
FROM (
SELECT mintime + INTERVAL seq.seq DAY AS orderdate
FROM (
SELECT CURDATE() - INTERVAL 1 MONTH AS mintime,
CURDATE() AS maxtime
FROM obs
) AS minmax
JOIN seq_0_to_999999 AS seq
ON seq.seq < TIMESTAMPDIFF(DAY,mintime,maxtime)
) AS alldates
LEFT JOIN (
SELECT COUNT(*) as Norders, DATE(date) AS orderdate
FROM ORDERS
WHERE date <= NOW()
AND date >= NOW() - INTERVAL 1 MONTH
GROUP BY DAY(date)
) AS orders ON alldates.orderdate = orders.orderdate
ORDER BY alldates.orderdate ASC
Notice that you need the LEFT JOIN so the rows in your output result set will be preserved even if there's no data in your ORDERS table.
Where do you get this sequence table seq_0_to_999999? You can make it like this.
DROP TABLE IF EXISTS seq_0_to_9;
CREATE TABLE seq_0_to_9 AS
SELECT 0 AS seq 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;
DROP VIEW IF EXISTS seq_0_to_999;
CREATE VIEW seq_0_to_999 AS (
SELECT (a.seq + 10 * (b.seq + 10 * c.seq)) AS seq
FROM seq_0_to_9 a
JOIN seq_0_to_9 b
JOIN seq_0_to_9 c
);
DROP VIEW IF EXISTS seq_0_to_999999;
CREATE VIEW seq_0_to_999999 AS (
SELECT (a.seq + (1000 * b.seq)) AS seq
FROM seq_0_to_999 a
JOIN seq_0_to_999 b
);
You can find an explanation of all this in more detail at http://www.plumislandmedia.net/mysql/filling-missing-data-sequences-cardinal-integers/
If you're using MariaDB version 10+, these sequence tables are built in.
First create a Calendar Table
SELECT coalesce(COUNT(O.*),0) as Norders, DATE_FORMAT(C.date, "%M %e") as sdate
FROM Calendar C
LEFT JOIN ORDERS O ON C.date=O.date
WHERE O.date <= NOW() AND O.date >= NOW() - INTERVAL 1 MONTH
GROUP BY DAY(date)
ORDER BY date ASC;
I have a table with sell orders and I want to list the COUNT of sell orders per day, between two dates, without leaving date gaps.
This is what I have currently:
SELECT COUNT(*) as Norders, DATE_FORMAT(date, "%M %e") as sdate
FROM ORDERS
WHERE date <= NOW()
AND date >= NOW() - INTERVAL 1 MONTH
GROUP BY DAY(date)
ORDER BY date ASC;
The result I'm getting is as follows:
6 May 1
14 May 4
1 May 5
8 Jun 2
5 Jun 15
But what I'd like to get is:
6 May 1
0 May 2
0 May 3
14 May 4
1 May 5
0 May 6
0 May 7
0 May 8
.....
0 Jun 1
8 Jun 2
.....
5 Jun 15
Is that possible?
Creating a range of dates on the fly and joining that against you orders table:-
SELECT sub1.sdate, COUNT(ORDERS.id) as Norders
FROM
(
SELECT DATE_FORMAT(DATE_SUB(NOW(), INTERVAL units.i + tens.i * 10 + hundreds.i * 100 DAY), "%M %e") as sdate
FROM (SELECT 0 i 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 i 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 i 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_SUB(NOW(), INTERVAL units.i + tens.i * 10 + hundreds.i * 100 DAY) BETWEEN DATE_SUB(NOW(), INTERVAL 1 MONTH) AND NOW()
) sub1
LEFT OUTER JOIN ORDERS
ON sub1.sdate = DATE_FORMAT(ORDERS.date, "%M %e")
GROUP BY sub1.sdate
This copes with date ranges of up to 1000 days.
Note that it could be made more efficient easily depending on the type of field you are using for your dates.
EDIT - as requested, to get the count of orders per month:-
SELECT aMonth, COUNT(ORDERS.id) as Norders
FROM
(
SELECT DATE_FORMAT(DATE_SUB(NOW(), INTERVAL months.i MONTH), "%Y%m") as sdate, DATE_FORMAT(DATE_SUB(NOW(), INTERVAL months.i MONTH), "%M") as aMonth
FROM (SELECT 0 i 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 UNION SELECT 10 UNION SELECT 11)months
WHERE DATE_SUB(NOW(), INTERVAL months.i MONTH) BETWEEN DATE_SUB(NOW(), INTERVAL 12 MONTH) AND NOW()
) sub1
LEFT OUTER JOIN ORDERS
ON sub1.sdate = DATE_FORMAT(ORDERS.date, "%Y%m")
GROUP BY aMonth
You are going to need to generate a virtual (or physical) table, containing every date in the range.
That can be done as follows, using a sequence table.
SELECT mintime + INTERVAL seq.seq DAY AS orderdate
FROM (
SELECT CURDATE() - INTERVAL 1 MONTH AS mintime,
CURDATE() AS maxtime
FROM obs
) AS minmax
JOIN seq_0_to_999999 AS seq ON seq.seq < TIMESTAMPDIFF(DAY,mintime,maxtime)
Then, you join this virtual table to your query, as follows.
SELECT IFNULL(orders.Norders,0) AS Norders, /* show zero instead of null*/
DATE_FORMAT(alldates.orderdate, "%M %e") as sdate
FROM (
SELECT mintime + INTERVAL seq.seq DAY AS orderdate
FROM (
SELECT CURDATE() - INTERVAL 1 MONTH AS mintime,
CURDATE() AS maxtime
FROM obs
) AS minmax
JOIN seq_0_to_999999 AS seq
ON seq.seq < TIMESTAMPDIFF(DAY,mintime,maxtime)
) AS alldates
LEFT JOIN (
SELECT COUNT(*) as Norders, DATE(date) AS orderdate
FROM ORDERS
WHERE date <= NOW()
AND date >= NOW() - INTERVAL 1 MONTH
GROUP BY DAY(date)
) AS orders ON alldates.orderdate = orders.orderdate
ORDER BY alldates.orderdate ASC
Notice that you need the LEFT JOIN so the rows in your output result set will be preserved even if there's no data in your ORDERS table.
Where do you get this sequence table seq_0_to_999999? You can make it like this.
DROP TABLE IF EXISTS seq_0_to_9;
CREATE TABLE seq_0_to_9 AS
SELECT 0 AS seq 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;
DROP VIEW IF EXISTS seq_0_to_999;
CREATE VIEW seq_0_to_999 AS (
SELECT (a.seq + 10 * (b.seq + 10 * c.seq)) AS seq
FROM seq_0_to_9 a
JOIN seq_0_to_9 b
JOIN seq_0_to_9 c
);
DROP VIEW IF EXISTS seq_0_to_999999;
CREATE VIEW seq_0_to_999999 AS (
SELECT (a.seq + (1000 * b.seq)) AS seq
FROM seq_0_to_999 a
JOIN seq_0_to_999 b
);
You can find an explanation of all this in more detail at http://www.plumislandmedia.net/mysql/filling-missing-data-sequences-cardinal-integers/
If you're using MariaDB version 10+, these sequence tables are built in.
First create a Calendar Table
SELECT coalesce(COUNT(O.*),0) as Norders, DATE_FORMAT(C.date, "%M %e") as sdate
FROM Calendar C
LEFT JOIN ORDERS O ON C.date=O.date
WHERE O.date <= NOW() AND O.date >= NOW() - INTERVAL 1 MONTH
GROUP BY DAY(date)
ORDER BY date ASC;
Can I list somehow the dates of last 30 days in MySQL? Not from a table!
For example I think about like this:
SELECT date WHERE date BETWEEN SUBDATE(NOW(), INTERVAL 30 DAY) AND NOW();
Is that possible?
I hacked this together from someone else's code, but it seems to work:
SELECT DATE_FORMAT(m1, '%d %b %Y')
FROM (
SELECT SUBDATE( NOW() , INTERVAL 30 DAY) + INTERVAL m DAY AS m1
FROM (
select #rownum:=#rownum+1 as m from
(select 1 union select 2 union select 3 union select 4) t1,
(select 1 union select 2 union select 3 union select 4) t2,
(select 1 union select 2 union select 3 union select 4) t3,
(select 1 union select 2 union select 3 union select 4) t4,
(select #rownum:=-1) t0
) d1
) d2
WHERE m1 <= now()
ORDER BY m1
The original code by valex is here:
How to get a list of months between two dates in mysql
You can do this the "explicit" way. That is, generate a series of numbers and calculate the date:
select date(date_sub(now(), interval n.n day) as thedate
from (select 1 as n union all
select 2 union all
. . .
select 30
) n
SELECT COUNT(*),Date(createDate) FROM EEC_Order WHERE createDate
BETWEEN DATE_SUB(CURDATE(),INTERVAL 7 DAY ) AND CURDATE()
group by Date(createDate) ;
COUNT(*) Date(createDate)
3 2013-09-08
1 2013-09-11
2 2013-09-12
Above query is fetching from table that means previous 7 days from current date(assume 2013-09-13) based on group by createdate. Results are showing 3 rows. While fetching time ,I want to add date and count in above RESULT only (not in DB) dummy date which are not there like following. Please help......How to write query for this...
COUNT(*) Date(createDate)
0 2013-09-06
0 2013-09-07
3 2013-09-08
0 2013-09-09
0 2013-09-10
1 2013-09-11
2 2013-09-12
Regards
sakir
esquareinfo
You need a list of the dates and a left outer join. Here is one way to formulate the query:
SELECT COUNT(o.createDate), dates.thedate
FROM (select DATE_SUB(CURDATE(), INTERVAL 7 DAY ) as thedate union all
select DATE_SUB(CURDATE(), INTERVAL 6 DAY ) as thedate union all
select DATE_SUB(CURDATE(), INTERVAL 5 DAY ) as thedate union all
select DATE_SUB(CURDATE(), INTERVAL 4 DAY ) as thedate union all
select DATE_SUB(CURDATE(), INTERVAL 3 DAY ) as thedate union all
select DATE_SUB(CURDATE(), INTERVAL 2 DAY ) as thedate union all
select DATE_SUB(CURDATE(), INTERVAL 1 DAY ) as thedate union all
select DATE_SUB(CURDATE(), INTERVAL 0 DAY ) as thedate
) dates left outer join
EEC_Order o
on Date(createDate) = dates.thedate
group by dates.thedate ;
EDIT:
Doing this dynamically in MySQL isn't so easy, unless you have a numbers or dates table somewhere. The following takes this approach, by first creating a list of 1000 numbers (with a where clause) and then doing the join and aggregation:
select DATE_SUB(CURDATE(), INTERVAL n.n DAY ), count(e.CreateDate)
from (select (n1*100 + n2*10 + n3) as n
from (select 0 as n 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
) n1 cross join
(select 0 as n 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
) n2 cross join
(select 0 as n 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
) n3
where (n1*100 + n2*10 + n3) <= 30
) n left outer join
EEC_Order o
on Date(createDate) = DATE_SUB(CURDATE(), INTERVAL n.n DAY )
group by DATE_SUB(CURDATE(), INTERVAL n.n DAY );
I have a database of users. I would like to create a graph based on userbase growth. The query I have now is:
SELECT DATE(datecreated), count(*) AS number FROM users
WHERE DATE(datecreated) > '2009-06-21' AND DATE(datecreated) <= DATE(NOW())
GROUP BY DATE(datecreated) ORDER BY datecreated ASC
This returns almost what I want. If we get 0 users one day, that day is not returned as a 0 value, it is just skipped and the next day that has at least one user is returned. How can I get something like (psuedo-response):
date1 5
date2 8
date3 0
date4 0
date5 9
etc...
where the dates with zero show up in sequential order with the rest of the dates?
Thanks!
I hope you will figure out the rest.
select * from (
select date_add('2003-01-01 00:00:00.000', INTERVAL n5.num*10000+n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) as date from
(select 0 as num
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) n1,
(select 0 as num
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) n2,
(select 0 as num
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) n3,
(select 0 as num
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) n4,
(select 0 as num
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) n5
) a
where date >'2011-01-02 00:00:00.000' and date < NOW()
order by date
With
select n3.num*100+n2.num*10+n1.num as date
you will get a column with numbers from 0 to max(n3)*100+max(n2)*10+max(n1)
Since here we have max n3 as 3, SELECT will return 399, plus 0 -> 400 records (dates in calendar).
You can tune your dynamic calendar by limiting it, for example, from min(date) you have to now().
This question asks the same thing I think. Generally the accepted answer seems to be that you either do it in your application logic (read in what you have into an array, then loop through the array and create the missing dates), or you use temporary tables filled with the dates you wish to join.
This is better to do as:
-- 7 Days:
set #n:=date(now() + interval 1 day);
SELECT qb.day_series as days , COALESCE(col_byte, 0) as Bytes from tbl1 qa
right join (
select (select #n:= #n - interval 1 day) day_series from tbl1 limit 7 ) as qb
on date(qa.Timestamp) = qb.day_series and
qa.Timestamp > DATE_SUB(curdate(), INTERVAL 7 day) order by qb.day_series asc
-- 30 Days:
set #n:=date(now() + interval 1 day);
SELECT qb.day_series as days , COALESCE(col_byte, 0) as Bytes from tbl1 qa
right join (
select (select #n:= #n - interval 1 day) day_series from tbl1 limit 30 ) as qb
on date(qa.Timestamp) = qb.day_series and
qa.Timestamp > DATE_SUB(curdate(), INTERVAL 30 day) order by qb.day_series asc;
or without variable like this:
SELECT qb.day_series as days , COALESCE(col_byte, 0) as Bytes from tbl1 qa
right join (
select curdate() - INTERVAL a.a day as day_series 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
) as a ) as qb
on date(qa.Timestamp) = qb.day_series and
qa.Timestamp > DATE_SUB(curdate(), INTERVAL 7 day) order by qb.day_series asc;
Do a right outer join to a table, call it tblCalendar, that is pre-populated with the dates you wish to report on. And join on the date field.
Paul
Query is:
SELECT qb.dy as yourday, COALESCE(count(yourcolumn), 0) as yourcount from yourtable qa
right join (
select curdate() as dy union
select DATE_SUB(curdate(), INTERVAL 1 day) as dy union
select DATE_SUB(curdate(), INTERVAL 2 day) as dy union
select DATE_SUB(curdate(), INTERVAL 3 day) as dy union
select DATE_SUB(curdate(), INTERVAL 4 day) as dy union
select DATE_SUB(curdate(), INTERVAL 5 day) as dy union
select DATE_SUB(curdate(), INTERVAL 6 day) as dy
) as qb
on qa.dates = qb.dy
and qa.dates > DATE_SUB(curdate(), INTERVAL 7 day)
order by qb.dy asc;
and the result is:
+------------+-----------+
| yourday | yourcount |
+------------+-----------+
| 2015-06-24 | 274339 |
| 2015-06-25 | 0 |
| 2015-06-26 | 0 |
| 2015-06-27 | 0 |
| 2015-06-28 | 134703 |
| 2015-06-29 | 87613 |
| 2015-06-30 | 0 |
+------------+-----------+
On further thought, something like this should be what you want:
CREATE TEMPORARY TABLE DateSummary1 ( datenew timestamp ) SELECT DISTINCT(DATE(datecreated)) as datenew FROM users;
CREATE TEMPORARY TABLE DateSummary2 ( datenew timestamp, number int ) SELECT DATE(datecreated) as datenew, count(*) AS number FROM users
WHERE DATE(datecreated) > '2009-06-21' AND DATE(datecreated) <= DATE(NOW())
GROUP BY DATE(datecreated) ORDER BY datecreated ASC;
SELECT ds1.datenew,ds2.number FROM DateSummary1 ds1 LEFT JOIN DateSummary2 ds2 on ds1.datenew=ds2.datenew;
This gives you all the dates in the first table, and the count summary data in the second table. You might need to replace ds2.number with IF(ISNULL(ds2.number),0,ds2.number) or something similar.