I am generating mysql query to show the number of orders received in current month group by days of month.
The table structure of mysql is as follow:
order_id date
======== ==========
1234 2012-07-02
1235 2012-07-02
1236 2012-07-04
1237 2012-07-07
1238 2012-07-08
Now I want it to return following results using mysql statement
count(order_id) day
=============== ===
0 01
2 02
0 03
1 04
0 05
0 06
1 07
1 08
So on and so forth till the end of the month 30/31 depends on the days in month.
Looking forward to your suggestions and help.
Thanks.
SELECT
count(order_id),
dates.dte
FROM
(
SELECT '2011-02-01' + INTERVAL a + b DAY dte
FROM
(SELECT 0 a UNION SELECT 1 a UNION SELECT 2 UNION SELECT 3
UNION SELECT 4 UNION SELECT 5 UNION SELECT 6 UNION SELECT 7
UNION SELECT 8 UNION SELECT 9 ) d,
(SELECT 0 b UNION SELECT 10 UNION SELECT 20
UNION SELECT 30 UNION SELECT 40) m
WHERE '2011-02-01' + INTERVAL a + b DAY < '2011-03-01'
ORDER BY a + b
) dates
LEFT JOIN
orders ON orders.`date` = dates.dte
GROUP BY
dates.dte
ORDER BY
dates.dte
generate day column values in your programming language
Thanks to #The Scrum Master for nice answer here
but I also agree this shouldn't be done in database. if you think about it you actually don't need rows with 0 count.
Prepare the calendar table with a list of months and days.
To retrieve all orders from 2012-07:
select day(c.date), count(*)
from calendar c
left join orders o on c.date=o.date
where year(c.date) = 2012
and month(c.date) = 07
group by day(c.date)
Related
I want to fetch data of last two months of each category from table.
Table looks like:
Id
Year
month
category
value
1
2019
1
TEST1
10
2
2018
12
TEST1
10
3
2018
10
TEST1
10
4
2018
1
TEST2
10
5
2018
12
TEST2
10
6
2018
1
TEST3
10
Expected output:
Id
Year
month
category
value
1
2019
1
TEST1
10
2
2018
12
TEST1
10
5
2018
12
TEST2
10
4
2018
1
TEST2
10
6
2018
1
TEST3
10
I tried using:
SELECT a.year,a.month,a.value, a.category
FROM test_data AS a
WHERE
(
SELECT COUNT(*)
FROM test_data AS b
WHERE b.category = a.category AND (b.year >= a.year AND b.month >= a.month)) <= 2
ORDER BY a.year DESC, a.month DESC
But it is giving extra record of TEST1 category. I guess because it is not working as expected for year condition. Please provide solution for this
Your first effort should go into fixing your data model and using a date-like datatype to store the date information, rather than spreading it over different columns. This should be as simple as adding a new column to the table and updating it from the existing columns:
alter table test_data add mydate date;
update test_data set mydate = concat(year, '-', month, '-01');
That said, to make your current query work you need to fine-tune the conditions on the dates as follows. One option uses arithmetics:
SELECT a.year, a.month, a.value, a.category
FROM test_data AS a
WHERE (
SELECT COUNT(*)
FROM test_data AS b
WHERE
b.category = a.category
AND 100 * b.year + b.month >= 100 * a.year + a.month
) <= 2
ORDER BY a.year DESC, a.month DESC
Alternatively, if you are running MySQL 8.0, you can use row_number():
SELECT a.year, a.month, a.value, a.category
FROM (
SELECT
a.*,
ROW_NUMBER() OVER(PARTITION BY a.category ORDER BY a.year DESC, a.month DESC)
FROM test_data AS a
) a
WHERE rn <= 2
I building an chart to display of Sales made in last 30 days.
I m getting output & other stuff for chart, but how do get dates for the label/legend/title? That too the date(label/legend title) changes everyday as i display data of last 30days. So i m looking to get output as dates of last 30 days.
If Current Date: 13-NOV-2019
Output example:
12-OCT-2019
13-OCT-2019
14-OCT-2019
...
...
...
12-NOV-2019
SELECT * FROM table WHERE start_date BETWEEN DATE(NOW()) AND DATE_SUB(DATE(NOW()), INTERVAL 30 DAY)
What you can do is
SELECT
DATE_FORMAT(#Date + Interval tr1.addd Day,'%d-%b-%Y')
FROM (SELECT #date := '2019-10-01') td
CROSS JOIN (SELECT 0 addd 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) tr1
This would give You
01-Oct-2019
02-Oct-2019
03-Oct-2019
04-Oct-2019
05-Oct-2019
06-Oct-2019
07-Oct-2019
08-Oct-2019
09-Oct-2019
10-Oct-2019
11-Oct-2019
With mysql 8.x
You can use CTE to generate the addays
WITH RECURSIVE cte AS (
SELECT 1 AS n
UNION ALL
SELECT n + 1 FROM cte WHERE n < 10
)
SELECT n FROM cte
Which would gove you
n
1
2
3
4
5
6
7
8
9
10
I have a mysql database having my client details, i want to show hourly visits, page views in a array. I made below given mysql query.
<?php
$siteid=$_GET["site"];
$DB = mysqli_connect('localhost','username','password','databasename');
$Q = "SELECT time, COUNT(*), pw FROM clientstats WHERE siteid='$siteid' AND time > UNIX_TIMESTAMP(DATE_SUB(NOW(), INTERVAL 26970 SECOND)) GROUP BY HOUR( FROM_UNIXTIME( `time`) ) ORDER BY time ASC";
$R = mysqli_query($DB,$Q);
//start loop
//while or foreach
while($row = mysqli_fetch_assoc($R)){
date_default_timezone_set('Asia/Kolkata');
$timestamp = $row['time'];
$timee = date('H', $timestamp);
$times = $row['COUNT(*)']+$row['wh'];
echo ",['$timee', {$row['COUNT(*)']}, $times]\r\n";
}
?>
where 26970 is seconds passed from midnight to till now. This code works fine and outputs a array like this
,['02', 2, 3] ,['02', 1, 2] ,['04', 1, 2] ,['05', 1, 2] ,['06', 2, 3]
This array structure is ['HOUR', visits, pw] but the problem is that this does not show values for each hour continuously like 1,2,3,4,5,6.... etc. I want to show values for each hours. If values for that hours are nil then i want to output that hour value as ['hour', 0, 0], Besides there is doubling in above output for second hour. How to correct it also.
Please help me, Thanks in advance.
To get the result returned from MySQL, using a SELECT statement, there needs to be a row source that returns the hour values that are "missing" from the table we are getting the count from.
(Getting this result from a query is only one option; we could approach it differently, and do it in the client. For example, populating an array with the hours we want, then run the original query, and assign the count to the appropriate matching array element. This answer is a demonstration of how we can get MySQL to return the specified result.)
As one example of how we can do that, here's a query that returns 24 rows of integer values 0 thru 23.
SELECT 00 AS hr UNION ALL SELECT 01 UNION ALL SELECT 02 UNION ALL SELECT 03
UNION ALL SELECT 04 UNION ALL SELECT 05 UNION ALL SELECT 06 UNION ALL SELECT 07
UNION ALL SELECT 08 UNION ALL SELECT 09 UNION ALL SELECT 10 UNION ALL SELECT 11
UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18 UNION ALL SELECT 19
UNION ALL SELECT 20 UNION ALL SELECT 21 UNION ALL SELECT 22 UNION ALL SELECT 23
We can wrap that query in parens, and reference that in place of a table in another query.
SELECT i.hr
FROM ( -- 0 thru 23
SELECT 00 AS hr UNION ALL SELECT 01 UNION ALL SELECT 02 UNION ALL SELECT 03
UNION ALL SELECT 04 UNION ALL SELECT 05 UNION ALL SELECT 06 UNION ALL SELECT 07
UNION ALL SELECT 08 UNION ALL SELECT 09 UNION ALL SELECT 10 UNION ALL SELECT 11
UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18 UNION ALL SELECT 19
UNION ALL SELECT 20 UNION ALL SELECT 21 UNION ALL SELECT 22 UNION ALL SELECT 23
) i
ORDER BY i.hr
There are other ways to return the set of integers. This example is just an example of an inline view (derived table), that is contained within the query. This doesn't depend on any additional tables in the database.
If we have a table in the database that contains 24 rows with integer values 0 thru 23, we could reference that in place of the inline view.
We can also add a WHERE clause to the outer query, to exclude rows that hvae an hr value is later than the current hour.
And if we want to return a two digit hour (with a leading zero), we can use an expression in SELECT list. There are several expressions that will do the trick.
SELECT RIGHT(CONCAT('0',i.hr),2) AS hh
FROM ( -- 0 thru 23
SELECT 00 AS hr UNION ALL SELECT 01 UNION ALL SELECT 02 UNION ALL SELECT 03
UNION ALL SELECT 04 UNION ALL SELECT 05 UNION ALL SELECT 06 UNION ALL SELECT 07
UNION ALL SELECT 08 UNION ALL SELECT 09 UNION ALL SELECT 10 UNION ALL SELECT 11
UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18 UNION ALL SELECT 19
UNION ALL SELECT 20 UNION ALL SELECT 21 UNION ALL SELECT 22 UNION ALL SELECT 23
) i
WHERE i.hr <= HOUR(NOW())
ORDER BY i.hr
We "match" the rows returned from that, with the counts returned from the original query. We can use an outer join operation so we return a row for an hour when there is no matching row from the count query.
To get a "missing" count returned as zero, we perform a conditional test, and replace a NULL value with a 0.
SELECT RIGHT(CONCAT('0',i.hr,2)) AS hh
, IFNULL(t.cnt,0) AS visits
, IFNULL(t.pw,0) AS pw
FROM ( -- 0 thru 23
SELECT 00 AS hr UNION ALL SELECT 01 UNION ALL SELECT 02 UNION ALL SELECT 03
UNION ALL SELECT 04 UNION ALL SELECT 05 UNION ALL SELECT 06 UNION ALL SELECT 07
UNION ALL SELECT 08 UNION ALL SELECT 09 UNION ALL SELECT 10 UNION ALL SELECT 11
UNION ALL SELECT 12 UNION ALL SELECT 13 UNION ALL SELECT 14 UNION ALL SELECT 15
UNION ALL SELECT 16 UNION ALL SELECT 17 UNION ALL SELECT 18 UNION ALL SELECT 19
UNION ALL SELECT 20 UNION ALL SELECT 21 UNION ALL SELECT 22 UNION ALL SELECT 23
) i
LEFT
JOIN ( -- counts by hour
SELECT HOUR(FROM_UNIXTIME(s.time)) AS hr
, COUNT(*) AS cnt
, MAX(s.pw) AS pw
FROM clientstats s
WHERE s.siteid = '$siteid'
AND s.time >= UNIX_TIMESTAMP(DATE(NOW()))
AND s.time < UNIX_TIMESTAMP(NOW())
GROUP BY HOUR(FROM_UNIXTIME(s.time))
) t
ON t.hr = i.hr
WHERE i.hr <= HOUR(NOW())
ORDER BY i.hr ASC
There are several different queries that will return the specified result. But the trick... to get zero counts returned, is to have a row source that returns the hour values we want returned, even when those hour values don't exist in the table we're getting a count from.
I have a table with 2 columns, date and score. It has at most 30 entries, for each of the last 30 days one.
date score
-----------------
1.8.2010 19
2.8.2010 21
4.8.2010 14
7.8.2010 10
10.8.2010 14
My problem is that some dates are missing - I want to see:
date score
-----------------
1.8.2010 19
2.8.2010 21
3.8.2010 0
4.8.2010 14
5.8.2010 0
6.8.2010 0
7.8.2010 10
...
What I need from the single query is to get: 19,21,9,14,0,0,10,0,0,14... That means that the missing dates are filled with 0.
I know how to get all the values and in server side language iterating through dates and missing the blanks. But is this possible to do in mysql, so that I sort the result by date and get the missing pieces.
EDIT: In this table there is another column named UserID, so I have 30.000 users and some of them have the score in this table. I delete the dates every day if date < 30 days ago because I need last 30 days score for each user. The reason is I am making a graph of the user activity over the last 30 days and to plot a chart I need the 30 values separated by comma. So I can say in query get me the USERID=10203 activity and the query would get me the 30 scores, one for each of the last 30 days. I hope I am more clear now.
MySQL doesn't have recursive functionality, so you're left with using the NUMBERS table trick -
Create a table that only holds incrementing numbers - easy to do using an auto_increment:
DROP TABLE IF EXISTS `example`.`numbers`;
CREATE TABLE `example`.`numbers` (
`id` int(10) unsigned NOT NULL auto_increment,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
Populate the table using:
INSERT INTO `example`.`numbers`
( `id` )
VALUES
( NULL )
...for as many values as you need.
Use DATE_ADD to construct a list of dates, increasing the days based on the NUMBERS.id value. Replace "2010-06-06" and "2010-06-14" with your respective start and end dates (but use the same format, YYYY-MM-DD) -
SELECT `x`.*
FROM (SELECT DATE_ADD('2010-06-06', INTERVAL `n`.`id` - 1 DAY)
FROM `numbers` `n`
WHERE DATE_ADD('2010-06-06', INTERVAL `n`.`id` -1 DAY) <= '2010-06-14' ) x
LEFT JOIN onto your table of data based on the time portion:
SELECT `x`.`ts` AS `timestamp`,
COALESCE(`y`.`score`, 0) AS `cnt`
FROM (SELECT DATE_FORMAT(DATE_ADD('2010-06-06', INTERVAL `n`.`id` - 1 DAY), '%m/%d/%Y') AS `ts`
FROM `numbers` `n`
WHERE DATE_ADD('2010-06-06', INTERVAL `n`.`id` - 1 DAY) <= '2010-06-14') x
LEFT JOIN TABLE `y` ON STR_TO_DATE(`y`.`date`, '%d.%m.%Y') = `x`.`ts`
If you want to maintain the date format, use the DATE_FORMAT function:
DATE_FORMAT(`x`.`ts`, '%d.%m.%Y') AS `timestamp`
I'm not a fan of the other answers, requiring tables to be created and such. This query does it efficiently without helper tables.
SELECT
IF(score IS NULL, 0, score) AS score,
b.Days AS date
FROM
(SELECT a.Days
FROM (
SELECT curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY AS Days
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.Days >= curdate() - INTERVAL 30 DAY) b
LEFT JOIN your_table
ON date = b.Days
ORDER BY b.Days;
So lets dissect this.
SELECT
IF(score IS NULL, 0, score) AS score,
b.Days AS date
The if will detect days that had no score and set them to 0. b.Days is the configured amount of days you chose to get from the current date, up to 1000.
(SELECT a.Days
FROM (
SELECT curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY AS Days
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.Days >= curdate() - INTERVAL 30 DAY) b
This subquery is something I saw on stackoverflow. It efficiently generates a list of the past 1000 days from the current date. The interval (currently 30) in the WHERE clause at the end determines which days are returned; the maximum is 1000. This query could be easily modified to return 100s of years worth of dates, but 1000 should be good for most things.
LEFT JOIN your_table
ON date = b.Days
ORDER BY b.Days;
This is the part that brings your table that contains the score into it. You compare to the selected date range from the date generator query to be able to fill in 0s where needed (the score will be set to NULL initially, because it is a LEFT JOIN; this is fixed in the select statement). I also order it by the dates, just because. This is preference, you could also order by score.
Before the ORDER BY you could easily join with your table about user info you mentioned with your edit, to add that last requirement.
I hope this version of the query helps someone. Thanks for reading.
Time went by since this question was asked. MySQL 8.0 was released in 2018 and added support for recursive common table expressions, which provide an elegant, state-of-the-art solution to this question.
The following query can be used to generate a list of dates, say for the first 15 days of August 2010:
with recursive all_dates(dt) as (
-- anchor
select '2010-08-01' dt
union all
-- recursion with stop condition
select dt + interval 1 day from all_dates where dt < '2010-08-15'
)
select * from all_dates order by dt
You can then left join this resultset with your table to generate the expected output:
with recursive all_dates(dt) as (
select '2010-08-01' dt
union all
select dt + interval 1 day from all_dates where dt < '2010-08-15'
)
select d.dt date, coalesce(t.score, 0) score
from all_dates d
left join mytable t on t.date = d.dt
order by d.dt
Demo on DB Fiddle:
date | score
:--------- | ----:
2010-08-01 | 19
2010-08-02 | 21
2010-08-03 | 0
2010-08-04 | 14
2010-08-05 | 0
2010-08-06 | 0
2010-08-07 | 10
2010-08-08 | 0
2010-08-09 | 0
2010-08-10 | 14
2010-08-11 | 0
2010-08-12 | 0
2010-08-13 | 0
2010-08-14 | 0
2010-08-15 | 0
Note that it is very easy to adapt the recursive CTE for other intervals or periods. As an example, say we want a row every 15 minutes from 4 AM to 8 AM on August 1st, 2010 ; we can do :
with recursive all_dates(dt) as (
select '2010-08-01 04:00:00' dt
union all
select dt + interval 15 minute from all_dates where dt < '2010-08-01 08:00:00'
)
...
You can accomplish this by using a Calendar Table. That's a table which you create once and fill with a date range (e.g. one dataset for each day 2000-2050; that depends on your data). Then you can make an outer join of your table against the calendar table. If a date is missing in your table, you return 0 for the score.
Michael Conard answer is great but I needed intervals of 15 minutes where the time must always start at the top of every 15th minute:
SELECT a.Days
FROM (
SELECT FROM_UNIXTIME( FLOOR( UNIX_TIMESTAMP() / (15 * 60) ) * (15 * 60)) - INTERVAL 15 * (a.a + (10 * b.a) + (100 * c.a)) MINUTE AS Days
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.Days >= curdate() - INTERVAL 30 DAY
This will set the current time to the previous round 15th minute:
FROM_UNIXTIME( FLOOR( UNIX_TIMESTAMP() / (15 * 60) ) * (15 * 60))
And this will remove time with a 15 minute step:
- INTERVAL 15 * (a.a + (10 * b.a) + (100 * c.a)) MINUTE
If there's a simpler way to do it, please let me know.
you can user direct from start date up to today with insertion
with recursive all_dates(dt) as (
-- anchor
select '2021-01-01' dt
union all
-- recursion with stop condition
INSERT IGNORE INTO mytable (date,score) VALUES (dt + interval 1 day ,0 ) where dt + interval 1 day <= curdate()
)
select * from all_dates
I have a MySQL-Table
id mydate content
----------------------------------
1 2015-06-20 some content
2 2015-06-20 some content
3 2015-06-22 some content
Now I want to count the entries for each day:
SELECT DATE(mydate) Date, COUNT(DISTINCT id) dayCount FROM mytable
GROUP BY DATE(mydate) HAVING dayCount > -1 ORDER BY DATE(mydate) DESC
This works for me, result:
2015-06-20 = 2
2015-06-22 = 1
How can I fetch days without any entries? In my example the result should be:
2015-06-19 = 0
2015-06-20 = 2
2015-06-21 = 0
2015-06-22 = 1
2015-06-23 = 0
Based on this:
<?php
$today = date("Y-m-d");
$mystartdate = date_create($today);
date_sub($mystartdate, date_interval_create_from_date_string('14 days'));
$mystartdate = date_format($mystartdate, 'Y-m-d');
?>
Finaly I want to output the counts of the last 14 days, also with "0-days". Hope you understand my problem.
For this you can create new table that holds the increment numbers, but it's not a great idea. However, if doing it this way, use this table to construct a list of dates using DATE_ADD.
LEFT JOIN onto your table of data based on the time portion to achieve your list of dates
for more info go through the link
MySQL how to fill missing dates in range?
try below-
SELECT a.date_field, COUNT(DISTINCT b.id) dayCount FROM
(SELECT date_field FROM
(
SELECT
MAKEDATE(YEAR(NOW()),1) +
INTERVAL (MONTH(NOW())-1) MONTH +
INTERVAL daynum DAY date_field
FROM
(
SELECT t*10+u daynum
FROM
(SELECT 0 t UNION SELECT 1 UNION SELECT 2 UNION SELECT 3) A,
(SELECT 0 u 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) B
ORDER BY daynum
) AA
) AAA
WHERE MONTH(date_field) = MONTH(NOW()) ) a
LEFT JOIN mytable b ON a.date_field=DATE(b.mydate)
GROUP BY a.date_field HAVING dayCount > -1 ORDER BY a.date_field DESC;