I have a database, and I would like to get data where the username is equal to x and I would also like to get data regardless of the user.
I have the two queries:
SELECT MONTH(date) as month, COUNT(*) as count
from DB_NAME
WHERE YEAR(date) = YEAR(CURDATE()) AND userid = 'userid'
GROUP BY month
SELECT MONTH(date) as month, COUNT(*) as count
from DB_NAME
WHERE YEAR(date) = YEAR(CURDATE())
GROUP BY month
Expected output:
+-------+------+-----+
| month | user | all |
+-------+------+-----+
| 1 | 100 | 200 |
| 2 | 90 | 150 |
+-------+------+-----+
I think some conditional aggregation is what you are after here:
SELECT
MONTH(date) as month,
COUNT(CASE WHEN userid = 'userid' THEN 1 END) as user,
COUNT(*) as count
from DB_NAME
WHERE YEAR(date) = YEAR(CURDATE())
GROUP BY month;
Related
I have a MySQL table which is as follows
------------------------------------------------
| id | user_id | day | value |
| INT(11) | INT(11) | INT(8) | DECIMAL(5,2) |
------------------------------------------------
Day is (Ymd) number like this: 20191012
I want to grab the SUM(value) for Today and the past 29 days separately (30 days, based on day column)
I thought of a loop like this
for($i=0;$i<=30;$i++)
{
$query = "SELECT SUM(value) FROM table WHERE day=".date('Ymd', strtotime("-{$i} days"));
}
But how can I achieve this more efficiently?
Any help is appreciated.
You could use Group By to fetch it with a single MySQL-Query
SELECT day, SUM(value) FROM table GROUP BY day WHERE day >= ".date('Ymd', strtotime("-29 days"));
It should work because your day field is a number and will increase everyday so you can just compare it with gte
Expected output will be
---------------------------
| day | SUM(value) |
| INT(8) | |
---------------------------
| 20191012 | 15.25 |
---------------------------
| 20191011 | 29.13 |
---------------------------
| ... | ... |
---------------------------
J4I: If there are days without values in your dataset, the missing "day-row" is also missing in the output so it could be that there are less than 30 result-rows.
This query will work for any version of MySQL. It creates a temporary numbers table with the numbers from 0-29, then uses that to compute a list of days from today to 29 days earlier. This is then LEFT JOINed to the data table to compute the sum per day, with 0 sums where there is no data for a given day:
SELECT d.day, COALESCE(SUM(value), 0)
FROM (SELECT DATE_FORMAT(CURDATE() -INTERVAL n10.n * 10 + n.n DAY, '%Y%m%d') AS day
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) n
CROSS JOIN (SELECT 0 AS n UNION ALL SELECT 1 UNION ALL SELECT 2) n10) d
LEFT JOIN `table` t ON t.day = d.day
GROUP BY d.day
Demo on dbfiddle
i have database in which user device id and device registerd date (cr_date) stored i want to fetch list of total count of registerd devices on same day with date wise
Here below display table structure
device_ID | cr_date
--------------------
1 | 2016-06-02 18:02:13
2 | 2016-06-02 18:03:58
3 | 2016-06-02 18:04:11
4 | 2016-06-03 18:04:33
5 | 2016-06-03 18:04:33
6 | 2016-06-04 18:04:44
and i want below result
count | date
---------------------------------
3 | 2016-06-02
2 | 2016-06-03
1 | 2016-06-04
i used below query but dosent get success
SELECT device_type, count(*) AS duplicate_count
FROM (
SELECT device_type FROM tbl_app_deviceregister
GROUP BY device_type HAVING COUNT(device_type) > 0
) AS t
Did you want this;)
SELECT count(1) as `count`, date_format(cr_date, '%Y-%m-%d') as `date`
FROM tbl_app_deviceregister
GROUP BY `date`
ORDER BY `count` DESC
You should GROUP BY date value, not by device:
SELECT CAST(cr_date AS DATE) AS `date`, COUNT(device_id)
FROM tbl_app_deviceregister
GROUP BY CAST(cr_date AS DATE)
Your query doesn't seem to reference the same table structure that you've described, but from what I understand, would this work?
SELECT count(id), date_registered FROM (
SELECT
id,
DATE(FROM_UNIXTIME(cr_date)) as date_registered
FROM
tbl_app_deviceregister
) A GROUP BY date_registered
you group it by device_type which is wrong. try to group by date.
select COUNT(*) as count,CONVERT(date,cr_date) as date
from tbl_app_deviceregister
group by CONVERT(date,cr_date)
I want to compute some stats (with a MySQL backend) sorted by date and with dynamic intervals (week, month, year).
Here is a little example :
Mysql table :
tracker_click
| ID | SITE_ID | CREATED_AT |
| ---- |---------| --------------------|
| 153 | 2 | 2013-07-22 15:43:25 |
| 154 | 2 | 2013-07-25 16:45:46 |
| 2501 | 2 | 2013-09-15 17:45:48 |
I want to get the total click number by SITE_ID by week for the last month with one query
And the same thing by month for the last year.
An example of what I want by week for the last month is :
| click number | SITE_ID | BEGIN_DATE | END_DATE |
|----------------|---------|----------------------|---------------------|
| 25 | 2 | 2013-07-01 00:00:00 | 2013-07-08 00:00:00 |
| 19 | 2 | 2013-08-09 00:00:00 | 2013-08-16 00:00:00 |
| 53 | 2 | 2013-0717- 00:00:00 | 2013-08-24 00:00:00 |
I don’t know if there is a solution to get exaclty this array with only one query without any other processes.
Thank you
This should get you the counts for the last month (ie, last 4 weeks), including weeks where the count is 0 for each site id. If you have a table of sites to get the site id from it means the cross join to the sub query can be replaced with a simple cross join to a table.
This generates a range of numbers from 0 to 5 and subtracts that number of weeks from the current date, formats that to give the Sunday and Saturday of the resulting week and checks that the resuling week is a week between the current date and the current date minus 1 month (done this way rather that just subtracting 4 weeks to cope with variable length months).
SELECT Weeks.aWeek_start, Weeks.aWeek_end, all_site_id.site_id, COUNT(tracker_click.id)
FROM
(
SELECT STR_TO_DATE(DATE_FORMAT(DATE_SUB(NOW(), INTERVAL units.i WEEK), '%Y%U Sunday 00:00:00'), '%X%V %W %H:%i:%s') AS aWeek_start,
STR_TO_DATE(DATE_FORMAT(DATE_SUB(NOW(), INTERVAL units.i WEEK), '%Y%U Saturday 23:59:59'), '%X%V %W %H:%i:%s') AS aWeek_end
FROM (SELECT 0 i UNION SELECT 1 UNION SELECT 2 UNION SELECT 3 UNION SELECT 4 UNION SELECT 5)units
WHERE DATE_FORMAT(DATE_SUB(NOW(), INTERVAL units.i WEEK), '%Y%U') BETWEEN DATE_FORMAT(DATE_SUB(NOW(), INTERVAL 1 MONTH), '%Y%U') AND DATE_FORMAT(NOW(), '%Y%U')
) Weeks
CROSS JOIN
(
SELECT DISTINCT site_id
FROM tracker_click
) AS all_site_id
LEFT OUTER JOIN tracker_click
ON tracker_click.CREATED_AT BETWEEN Weeks.aWeek_start AND Weeks.aWeek_end
AND tracker_click.site_id = all_site_id.site_id
GROUP BY Weeks.aWeek_start, Weeks.aWeek_end, all_site_id.site_id
A similar query could be done for months of the year
SELECT Months.aMonth_start, Months.aMonth_end, all_site_id.site_id, COUNT(tracker_click.id)
FROM
(
SELECT DATE_FORMAT(DATE_SUB(NOW(), INTERVAL units.i MONTH), '%Y/%m/01 00:00:00') AS aMonth_start,
DATE_FORMAT(LAST_DAY(DATE_SUB(NOW(), INTERVAL units.i MONTH)), '%Y/%m/%d 23:59:59') AS aMonth_end
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)units
) Months
CROSS JOIN
(
SELECT DISTINCT site_id
FROM tracker_click
) AS all_site_id
LEFT OUTER JOIN tracker_click
ON tracker_click.CREATED_AT BETWEEN Months.aMonth_start AND Months.aMonth_end
AND tracker_click.site_id = all_site_id.site_id
GROUP BY Months.aMonth_start, Months.aMonth_end, all_site_id.site_id
SELECT count(*), site_id, extract(WEEK from created_at) as start_date,
date_add(extract(WEEK from created_at), 1, week) as end_date
FROM click_tracker
GROUP BY site_id, extract(WEEK from created_at)
and you can add where clause to filter results additionally.
PaymentHistory Table
==========================
UserId | Amount | Date
==========================
1 | 300 | 2014-06-26
3 | 300 | 2014-06-26
2 | 200 | 2014-06-26
1 | 400 | 2014-06-22
1 | 100 | 2014-06-21
display two columns first is sum of amount as Named "currentEarning" which has same date of today and second is sum of amount as Named "oldEarning" based on previous date of today. (for particular userId) see output:
output
=====================================
UserId | currentEarning | oldEarning |
=====================================
1 | 300 | 500
You can do that with one query:
SELECT
SUM(IF(`Date` = CURRENT_DATE, AMOUNT, 0)) AS currentEarning,
SUM(IF(`Date` < CURRENT_DATE, AMOUNT, 0)) AS oldEarning
FROM PaymentHistory
GROUP BY `UserId`
Way more efficient than subqueries or two separate queries.
Get current amount:
SELECT UserId, SUM(Amount) as currentEarning, Date FROM PaymentHistory WHERE UserId = '1' AND DATE(Date) = CURDATE()
Get old amount:
SELECT UserId, SUM(Amount) as oldEarning, Date FROM PaymentHistory WHERE UserId = '1' AND DATE(Date) != CURDATE()
This should work
SELECT `UserId`,
(SELECT SUM(`Amount`) FROM `PaymentHistory` WHERE `UserID` = t1.`UserId` AND `Date` = t1.`Date`) AS `currentEarning`,
(SELECT SUM(`Amount`) FROM `PaymentHistory` WHERE `UserID` = t1.`UserId` AND `Date` < t1.`Date`) AS `oldEarning`
FROM PaymentHistory AS t1
WHERE t1.`Date` = CURRENT_DATE
GROUP BY t1.`UserId`
HAVING `oldEarning` IS NOT NULL
I have a database with the following format:
myTable
productgroupID | productID | views | date
1 | 10 | 25 | 2013-05-23
4 | 105 | 15 | 2013-05-23
7 | 60 | 65 | 2013-05-23
7 | 60 | 55 | 2013-05-22
7 | 60 | 45 | 2013-05-21
Now I want to sum all views of a product in the moth May.
Result should be:
productgroupID | productID | viewed | month
7 | 60 | 165 | 2013-05-01
1 | 10 | 25 | 2013-05-01
4 | 105 | 15 | 2013-05-01
I tried the query below, but this gives me all views of a specific productgroupID. But I need the sum of the unique productgroupID & productID.
SELECT COUNT( views ) AS viewed, productgroupID FROM product_stats_daily GROUP BY productgroupID
If you want the views totaled, then you can use the sum() aggregate function and then you can group by the month and year for the date:
select productGroupId,
productId,
sum(views) viewed,
month(date) Month,
year(date) Year
from myTable
group by productGroupId, productId, month(date), year(date);
See SQL Fiddle with Demo
You could also use Date_Format to get the date in the format that you want:
select productGroupId,
productId,
sum(views) viewed,
DATE_FORMAT(date, '%Y-%m-01') date
from myTable
group by productGroupId, productId, DATE_FORMAT(date, '%Y-%m-01')
See SQL Fiddle with Demo
COUNT will count number of rows while SUM will sum up value of retrieved rows.
So your query becomes:
SELECT SUM( views ) AS viewed, productgroupID
FROM product_stats_daily
GROUP BY productgroupID
There's probably a more elegant way to force the date to the beginning of the month, but this should work:
SELECT
ProductGroupID,
ProductID,
SUM(views) AS viewed,
DATE_FORMAT(date, '%Y-%m-01') AS Month
FROM myTable
GROUP BY
ProductGroupID,
ProductID,
DATE_FORMAT(date, '%Y-%m-01')