For example I have below table:
---------------------
| amount | date |
---------------------
| 50 | Day 1 |
| 60 | day 2 |
| 20 | day 3 |
| 25 | day 3 |
| 23 | day 4 |
| 26 | day 4 |
| 15 | day 5 |
---------------------
What I basically want to do is to retrieve item from last 2 days and order the row by max amount. so the result would be like:
---------------------
| amount | date |
---------------------
| 26 | day 4 |
| 23 | day 4 |
| 15 | day 5 |
---------------------
FYI: the date input is in 2018-06-10 14:37:44 mode above is just an example
I tried: SELECT AMOUNT FROM table WHERE AMOUNT=(SELECT MAX(AMOUNT) FROM table) ORDER BY DATE;
But the result I am getting is only | 60 | day 2 | which is the max amount and is not from 2 recent dates;
You could use the DATE_ADD function that gives you a date after a certain time alongside CURDATE() that gives you current date;
SELECT * FROM table WHERE date >= DATE_ADD(CURDATE(), INTERVAL -2 DAY)
ORDER BY amount DESC;
Try this:
SELECT amount, date FROM table WHERE date >= DATE_ADD(CURDATE(), INTERVAL -2 DAY) ORDER BY amount DESC;
-- date > DATE_ADD(CURDATE(), INTERVAL -2 DAY) will give you all rows within 2 days
-- ORDER BY amount DESC will give you ordered by amount descending
Related
Here is a part of table from which I am retrieving data for the last 3 months including current month
+-------------+-----------------------+
| Wo_id | updated_at |
+-------------+-----------------------+
| 1 | 2018-12-05 10:38:06 |
| 2 | 2018-12-02 15:21:17 |
| 3 | 2018-12-01 22:18:53 |
| 4 | 2018-10-25 10:38:06 |
| 5 | 2018-10-18 15:21:17 |
| 6 | 2018-10-16 22:18:53 |
| 7 | 2018-10-19 10:26:19 |
| 8 | 2018-10-27 07:06:52 |
| 9 | 2018-09-25 11:35:09 |
| 10 | 2018-09-18 12:54:27 |
The query I tried is
SELECT MONTHNAME(updated_at) month,YEAR(updated_at) year_name,
MONTH(updated_at) month_no, COUNT(*) work_orders
FROM work_orders where updated_at >= last_day(now()) + interval 1 day - interval 3 month
GROUP by MONTH(updated_at),YEAR(updated_at)
ORDER BY MONTH(updated_at) DESC
The Output I am getting is
+-------------+-------------+----------+-------------+
| month | year_name | month_no | work_orders |
+-------------+-------------+----------+-------------+
| December | 2018 | 12 | 3 |
| October | 2018 | 10 | 5 |
| September | 2018 | 9 | 2 |
As you can see the query is neglecting November as its data is not in the table. It is Including September in order to complete the cycle of 3 months which is wrong. I want the output like this
+-------------+-------------+----------+-------------+
| month | year_name | month_no | work_orders |
+-------------+-------------+----------+-------------+
| December | 2018 | 12 | 3 |
| November | 2018 | 9 | 0 |
| October | 2018 | 10 | 5 |
Can someone guide me in modifying the above mentioned query. Thanks
You need to create a table of the last three months and then LEFT JOIN that to your work orders table (using the month of the work order) to get the results you want. The table of the last 3 months can be generated using a UNION:
SELECT NOW() AS month
UNION
SELECT NOW() - INTERVAL 1 MONTH
UNION
SELECT NOW() - INTERVAL 2 MONTH
Output (as of 2018-12-07):
month
2018-12-07 11:06:15
2018-11-07 11:06:15
2018-10-07 11:06:15
Note that it is OK to subtract 1 month from the date as if the day number is larger than the number of days in the previous month it will be adjusted downward to make the date valid (see the manual).
The final query then becomes:
SELECT MONTHNAME(m.month) AS month_name, YEAR(m.month) AS year_name,
MONTH(m.month) AS month_no, COUNT(wo.Wo_id) work_orders
FROM (SELECT NOW() AS month
UNION
SELECT NOW() - INTERVAL 1 MONTH
UNION
SELECT NOW() - INTERVAL 2 MONTH) m
LEFT JOIN work_orders wo ON MONTH(wo.updated_at) = MONTH(m.month) AND
YEAR(wo.updated_at) = YEAR(m.month)
GROUP by m.month, year_name
ORDER BY m.month DESC
Note that we don't need a WHERE clause as the values in the month table restrict the data to the last 3 months that we are interested in. Also we use a LEFT JOIN so that we get a result for each month even if there were no work orders that month.
Output:
month_name year_name month_no work_orders
December 2018 12 3
November 2018 11 0
October 2018 10 5
Demo on dbfiddle
Storage table
| id| product_id | date_add | date_remove
------------------------------------------------------------------
| 1 | 10 |2018-04-02 08:28:43 | 2018-04-03 07:21:08
| 2 | 10 |2018-04-05 08:28:43 | 2018-04-06 08:28:50
| 3 | 10 |2018-04-01 08:28:43 | 2018-04-05 08:28:50
| 4 | 12 |2018-04-01 08:28:43 | 2018-04-03 07:21:08
| 5 | 12 |2018-04-04 08:28:43 | 2018-04-04 10:28:43
| 6 | 13 |2018-03-01 08:28:43 | 2018-03-01 10:28:43
how to find ?
how many days product was in the storage in period 2018-04-01 to 2018-04-05?
find result
| product_id | days
| 10 | 5
| 12 | 3
try
SELECT product_id, SUM(DATEDIFF(date_remove, date_add)) as days
FROM storage
where date_remove BETWEEN '2018-04-01 00:00:00'
AND '2018-04-05 23:59:59'
AND date_add BETWEEN '2018-04-01 00:00:00'
AND '2018-04-05 23:59:59'
GROUP BY product_id
but result wrong because 'SUM' sums all days
get result
| product_id | days
| 10 | 7
correct result
| product_id | days
| 10 | 5
upd
http://rextester.com/QFS96125
result 9,646805555556 but probably maximum 5 days and product_id 13 correct 0,436608796296 but result 0,87
First you want to look at all date ranges that are within or overlap with the range 2018-04-01 to 2018-04-05.
where date_add < date '2018-04-06' and date_remove >= date '2018-04-01'
Then, with the ranges found, you want to consider only their days in the range 2018-04-01 to 2018-04-05.
greatest(date_add, date '2018-04-01')
least(date_remove, date '2018-04-06')
Then you want to count days. Here you need a rule. Do you want to look at single ranges and only take their full days which you add up then? Or do you want to consider day fractions, add all up and see how many full days result? For the latter you could get durations in seconds and add these up:
select
product_id,
sum(timestampdiff(second, greatest(date_add, date '2018-04-01'),
least(date_remove, date '2018-04-06'))
) / 60 / 60 / 24 as days
from storage
where date_add < date '2018-04-06' and date_remove >= date '2018-04-01'
group by product_id
order by product_id;
This gets you
product_id | days
-----------+---------------
10 | 5,599872685185
12 | 2,036400462963
Feel free to use FLOOR, CEIL or ROUND on the resulting days.
Rextester demo: http://rextester.com/XTVU47656
To obtain such result, try
SELECT SUM(DATEDIFF(date_remove, date_add)) as days
FROM table
GROUP BY product_id
Keep in mind that this will sum all the days for the same product_id. To get the result for each id, use:
SELECT id, product_id, DATEDIFF(date_remove, date_add) as days
FROM table
I know it's possible to get yesterday records, most common way using SUBDATE(CURDATE(), 1) or maybe simply use CURDATE() - 1 and use LIMIT and ORDER to retrieve the last record of yesterday.
But here, I need to get the last record of yesterday in the first row and the rest will be all records of today. I need to run this within single query.
For example, I have following records in one of my table:
--------------------------------------------------
| value | created_at |
--------------------------------------------------
| 70 | 1/1/2017 |
| 300 | 1/1/2017 |
| 100 | 1/1/2017 |
| 235 | 1/2/2017 |
| 45 | 1/2/2017 |
--------------------------------------------------
The created_at column is a timestamp, if today is 1/2/2017 (2th January 2017) then the result of the query should be:
--------------------------------------------------
| value | created_at |
--------------------------------------------------
| 100 | 1/1/2017 |
| 235 | 1/2/2017 |
| 45 | 1/2/2017 |
--------------------------------------------------
So far, I only able to retrieve the records of today with following query:
SELECT * FROM my_table WHERE created_at >= CURDATE();
What query I need to accomplish this?
Hoping you have id as primary key
select * from
(select
*
from
tbl
where date(created_at) =date(DATE_ADD(now(), INTERVAL -1 DAY))
order by id desc limit 0,1
)tmp
UNION
select * from tbl where date(created_at)=date(now())
I have the following table :
+----+-------------+
| ID | startDate |
+----+-------------+
| 1 | 2014-10-29 |
| 2 | 2014-10-29 |
| 3 | 2014-10-28 |
| 4 | 2014-10-28 |
| 5 | 2014-10-28 |
| 6 | 2014-10-26 |
| 7 | 2014-10-25 |
| 8 | 2014-10-23 |
| 9 | 2014-10-22 |
+----+-------------+
From this table, I want to get the last 5th day's startDate from current date. The date may not be consecutive I mean in this table my data will not enter everyday but whenever I want to check last 5th day from today it should return like '2014-10-23'. Please help me out.
Use Limit
SELECT startDate FROM table WHERE startDate <= CURDATE() ORDER BY startDate LIMIT 5,1
It means
1)Get records Whose start date is less than current date, and then
2) return one record starting at record 5(means get 5th record).
your query will look something like this
select startDate
from table_name
where startDate <= CURDATE()
order by startDate desc limit 5,1
or
select startDate
from table_name
where startDate <= CURDATE()
order by startDate desc limit 1 offset 5
I'm using MySQL and I have the following table:
| clicks | int |
| period | date |
I want to be able to generate reports like this, where periods are done in the last 4 weeks:
| period | clicks |
| 1/7 - 7/5 | 1000 |
| 25/6 - 31/7 | .... |
| 18/6 - 24/6 | .... |
| 12/6 - 18/6 | .... |
or in the last 3 months:
| period | clicks |
| July | .... |
| June | .... |
| April | .... |
Any ideas how to make select queries that can generate the equivalent date range and clicks count?
SELECT
WEEKOFYEAR(`date`) AS period,
SUM(clicks) AS clicks
FROM `tablename`
WHERE `date` >= CURDATE() - INTERVAL 4 WEEK
GROUP BY period
SELECT
MONTH(`date`) AS period,
SUM(clicks) AS clicks
FROM `tablename`
WHERE `date` >= CURDATE() - INTERVAL 3 MONTH
GROUP BY period
For the last 3 months you can use:
SELECT MONTH(PERIOD), SUM(CLICKS)
FROM TABLE
WHERE PERIOD >= NOW() - INTERVAL 3 MONTH
GROUP BY MONTH(PERIOD)
or for the last 4 weeks:
SELECT WEEK(PERIOD), SUM(CLICKS)
FROM TABLE
WHERE PERIOD >= NOW() - INTERVAL 4 WEEK
GROUP BY WEEK(PERIOD)
Code not tested.