I have table called room_prices that store different prices according to the date specified. Here's the sample data.
id room_id date_from date_to price
1 1 3500.00
25 1 7/28/2012 7/29/2012 3600.00
27 1 8/24/2012 8/27/2012 3000.00
49 1 11/12/2012 11/16/2012 4000.00
The first row is the default price of room 1 (room_id = 1).
When guest reserve a room, the system will first look at the prices on room_prices table. If the date reserved by guest was found on the table it will get the price according to the date selected and not the default price.
Ex.
If the guest reserved a room from 11/10/2012 to 11/11/2012 then the price should be 3,500.
If the guest reserved a room from 11/12/2012 to 11/14/2012 then the price should be 4,000.
If the guest reserved a room from 11/10/2012 to 11/13/2012 then the average price should be 3750. See table below:
11/10/2012 3500
11/11/2012 3500
11/12/2012 4000
11/13/2012 4000
3750
So my question is if there is a single SQL code to get the average price.
update:
I tried this sql to no avail:
SELECT avg(price) as avg_price
FROM lf_rooms_prices
WHERE room_id = 1
AND ((DATE_FORMAT(date_from, '%m/%d/%Y') >= '11/10/2012'
AND DATE_FORMAT(date_from, '%m/%d/%Y') < '11/10/2012')
OR (DATE_FORMAT(date_to, '%m/%d/%Y') > '11/13/2012'
AND DATE_FORMAT(date_to, '%m/%d/%Y') <= '11/13/2012'))
btw, I'm using codeigniter.
There's AVG() function, so to get average price your query should look like this:
SELECT AVG(`price`) FROM `room_prices` GROUP BY `room_id`
See docs here. And of course you can add WHERE to the query to narrow data set you want the average value of:
SELECT AVG(`price`) FROM `room_prices` WHERE <CONDITIONS> GROUP BY `room_id`
EDIT:
Full query:
SELECT AVG(`price`) FROM `room_prices`
WHERE `room_id` = 1
AND ((DATE_FORMAT(date_from, '%m/%d/%Y') >= '11/10/2012'
AND DATE_FORMAT(date_from, '%m/%d/%Y') < '11/10/2012')
OR (DATE_FORMAT(date_to, '%m/%d/%Y') > '11/13/2012'
AND DATE_FORMAT(date_to, '%m/%d/%Y') <= '11/13/2012'))
GROUP BY `room_id`
The key is to use GROUP BY.
Related
I have the following table with dummy values in mysql database:
id cnic amount depositDate receivedBy receivingZone remarks
1 11111 10000 01-Nov-2019 11111 1 Ok
2 11111 10000 07-Nov-2019 11111 1 ok
Now i want to get the sum of amount from the table for specific year (2019 in this case) where the year came from current timestamp (it may be 2020, 2021 etc depends on the current date)
Any help plz
You can use the YEAR function to get the year of the depositDate column and also the current year and then sum only the values which match:
SELECT SUM(amount) AS amount
FROM yourtable
WHERE YEAR(STR_TO_DATE(depositDate, '%d-%b-%Y')) = YEAR(CURDATE())
You can try below -
select sum(amount)
from tablename
where year(depositdate)=year(now())
I would write the WHERE clause to be sargable:
SELECT SUM(amount)
FROM yourTable
WHERE depositDate >= DATE_FORMAT(NOW() ,'%Y-01-01') AND
depositDate < DATE_FORMAT(DATE_ADD(NOW(), INTERVAL 1 YEAR) ,'%Y-01-01');
This approach, while a bit more verbose than the other answers which use the YEAR() function, would allow an index on the depositDate column to be used.
Based on your sample year, we need to recognize first the date using str_to_date
select sum(amount)
from tableA
where year(now()) = year(str_to_date(depositdate, '%d-%b-%Y'))
Am trying to find the min value from past 30 days, in my table there is one entry for every day, am using this query
SELECT MIN(low), date, low
FROM historical_data
WHERE name = 'bitcoin'
ORDER BY STR_TO_DATE(date,'%d-%m-%Y') DESC
LIMIT 7
But this value not returing the correct value. The structure of my table is
Table structure
And table data which is store is like this
Table data style
Now what i need is to get the minimum low value. But my query not working it give me wrong value which even did not exist in table as well.
Updates:
Here is my updated Table Structure.
enter image description here
And here is my data in this table which look like this
enter image description here
Now if you look at the data, i want to check the name of token omisego and fatch the low value from past 7 days which will be from 2017-12-25 to 2017-12-19
and in this cast the low value is 9.67, but my current query and the query suggested by my some member did not brings the right answer.
Update 2:
http://rextester.com/TDBSV28042
Here it is, basically i have more then 1400 coins and token historical data, which means that there will me more then 1400 entries for same date like 2017-12-25 but having different name, total i have more then 650000 records. so every date have many entries with different names.
To get the lowest row per group you could use following
SELECT a.*
FROM historical_data a
LEFT JOIN historical_data b ON a.name = b.name
AND a.low > b.low
WHERE b.name IS NULL
AND DATE(a.date) >= '2017-12-19' AND DATE(a.date) <= '2017-12-25'
AND a.name = 'omisego'
or
SELECT a.*
FROM historical_data a
JOIN (
SELECT name,MIN(low) low
FROM historical_data
GROUP BY name
) b USING(name,low)
WHERE DATE(a.date) >= '2017-12-19' AND DATE(a.date) <= '2017-12-25'
AND a.name = 'omisego'
DEMO
For last 30 day of 7 days or n days you could write above query as
SELECT a.*, DATE(a.`date`)
FROM historical_data2 a
LEFT JOIN historical_data2 b ON a.name = b.name
AND DATE(b.`date`) >= CURRENT_DATE() - INTERVAL 30 DAY
AND DATE(b.`date`) <= CURRENT_DATE()
AND a.low > b.low
WHERE b.name IS NULL
AND a.name = 'omisego'
AND DATE(a.`date`) >= CURRENT_DATE() - INTERVAL 30 DAY
AND DATE(a.`date`) <= CURRENT_DATE()
;
DEMO
But note it may return more than one records where low value is same, to choose 1 row among these you have specify another criteria to on different attribute
Consider grouping the same and running the clauses
SELECT name, date, MIN(low)
FROM historical_data
GROUP BY name
HAVING name = 'bitcoin'
AND STR_TO_DATE(date, '%M %d,%Y') > DATE_SUB(NOW(), INTERVAL 30 DAY);
Given the structure, the above query should get you your results.
// Try this code ..
SELECT MIN(`date`) AS date1,low
FROM historical_data
WHERE `date` BETWEEN now() - interval 1 month
AND now() ORDER by low ASC;
I have a table which has product quantity column and the database has multiple entries in a single month.
Date format is (YYYY-MM-DD)
Date Quantity
2016-03-01 1200
2016-03-05 200
2016-04-05 500
2016-04-10 1000
2016-05-05 850
2016-05-10 50
So I want data as:
March (2016-03-01 to 2016-03-31) = 1400
April (2016-04-01 to 2016-04-30) = 1500
May (2016-05-01 to 2016-05-31) = 900
How can I do this?
Use the following query and it will return the result that you want.
' date field ' is the name of the column where date is inserted and Tablename is the name of the Table.
SELECT MONTH(date field) as month , YEAR(date field) as year , SUM(quantity) as
quantity FROM Tablename GROUP BY MONTH( date field )
select MONTH(Date) As dt , SUM(quantity) total from #Table1
GROUP BY MONTH(Date)
You can use DATE_FORMAT to convert the date field into just the year-month format, select the SUM of the quantity field, and then GROUP_BY the year-month field.
SELECT DATE_FORMAT('%Y-%m', `date`) as `ym`, SUM(`quantity`) FROM `table` GROUP BY `ym` ORDER BY `ym` ASC
You can try grouping by the month and year, taking the sum of the quantity field as an aggregate.
SELECT MONTH(Date), YEAR(Date), SUM(quantity)
FROM yourTable
GROUP BY MONTH(Date), YEAR(Date)
If you want the fancy output with actual date ranges, that would be a bit more work to do. Those dates may not exist in your original data set, so it could require a date table. And handling February in a leap year could be a real pain.
Use the following query same as it is it will return what you want
SELECT CONCAT(DATE_FORMAT(date,'%b'),' ',
(SELECT CONCAT('(',YEAR(date),'-',LPAD(MONTH(date),2,0),'-01 - ')
), (SELECT concat(last_day(date),')')
)) dates, SUM(quantity) qty FROM DATES GROUP BY MONTH(date) ORDER BY MONTH(date)
I have 2 tables,
prices (pcode, date, priceperweek)
1 record covers 1 week
booked (pcode, date)
1 record = 1 day, because shortbreaks are available
And a form consists of 3 fields
startdate
flexibility (+/- 1/2/3 weeks)
duration (3 / 4 / 5 / 6 / 7 / 14)).
The below query should be selecting prices & dates and checking to see if the selected start date appears in the prices table and that each day from the startdate doesn't appear in the bookings table.
SELECT SUM(priceperweek) AS `ppw`, prices.date AS `startdate`
FROM `prices` LEFT JOIN `booked` ON prices.pcode=booked.pcode
WHERE prices.pcode='A2CD59GH'
AND (prices.date IN ('20131221', '20131228')
AND booked.date NOT IN ('20131221', '20131222', '20131223',
'20131224', '20131225', '20131226', '20131227', '20131228',
'20131229', '20131230', '20131231', '20140101', '20140102',
'20140103')
)
OR (prices.date IN ('20131214', '20131221')
AND booked.date NOT IN ('20131214', '20131215', '20131216',
'20131217',
'20131218', '20131219', '20131220', '20131221', '20131222',
'20131223', '20131224', '20131225', '20131226', '20131227')
)
OR (prices.date IN ('20131228', '20140104') AND booked.date NOT IN
('20131228', '20131229', '20131230', '20131231', '20140101',
'20140102', '20140103', '20140104', '20140105', '20140106',
'20140107', '20140108', '20140109', '20140110')
)
GROUP BY prices.date
ORDER BY prices.date ASC
VALUES GIVEN TO QUERY...
startdate = 20131221
duration = 14
property = A2CD59GH
plusminus = 1
My problem is that this query returns records even if some of the dates in a range appear in the "bookings" table AND the ppw value is alot more than i would have expected.
The reason for using SUM(ppw) is when a duration of 14 is specified the price will sum both weeks together.
Thanks for any help on this
The problem with your approach is that the startdate will only be filtered from the results if every single record in the booked table for the given pcode falls within the booking period. Obviously this won't be the case if the property has been booked on some other date.
I'd suggest performing an anti-join along the following lines:
SELECT t.date, SUM(prices.priceperweek) FROM prices JOIN (
SELECT prices.date
FROM prices LEFT JOIN booked
ON booked.pcode = prices.pcode
AND booked.date BETWEEN prices.date
AND prices.date + INTERVAL 14 DAY
WHERE booked.pcode IS NULL
AND prices.pcode = 'A2CD59GH'
AND prices.date BETWEEN '20131221' - INTERVAL 1 WEEK
AND '20131221' + INTERVAL 2 WEEK
) t ON prices.date BETWEEN t.date AND t.date + INTERVAL 13 DAY
GROUP BY t.date
I have a table revenue with following fields
+----+-----------+-------------+---------------+
| id | member_id | amount_paid | datetime_paid |
+----+-----------+-------------+---------------+
I want to run a query to find the top 10 maximum revenue days. What would be the fastest way to achieve this? The table has a lot of data.
Regards,
Try the basic query which does it:
SELECT DATE(`datetime_paid`) AS `Max Revenue`
FROM `revenue`
ORDER BY `amount_paid` DESC
GROUP BY DATE(`datetime_paid`)
LIMIT 10;
This shows you the Top 10 Max paid days. :)
Assuming your datetime_paid column is of some date or time type (see MySQL docu), the following should work.
SELECT SUM( `amount_paid` ) AS `amount`, DATE( `datetime_paid` ) as `date`
FROM yourTable
GROUP BY `date`
ORDER BY `amount` DESC
LIMIT 10
First you need to group the results by the day the payment was made. This can be done by selecting DATE(datetime_paid) as payday, then grouping by payday. To get the total, select SUM(amount_paid) as total. Finally, to get the top 10, order by total desc and limit 10.
Your final query should look like:
SELECT DATE(`datetime_paid`) AS `payday`, SUM(`amount_paid`) AS `total`
FROM `table_name_here`
GROUP BY `payday`
ORDER BY `total` DESC
LIMIT 10
You may need to change that ORDER BY line to ORDER BY SUM(amount_paid) DESC, I can't remember if field aliases are allowed in ordering.