calculate price between given dates - php

Hotel_id Room_id Room_type Start_date End_date Price
----------------------------------------------------------------
13 2 standard 2012-08-01 2012-08-15 7000
13 2 standard 2012-08-16 2012-08-31 7500
13 2 standard 2012-09-01 2012-09-30 6000
13 3 luxury 2012-08-01 2012-08-15 9000
13 3 luxury 2012-08-16 2012-08-31 10000
13 3 luxury 2012-09-01 2012-09-30 9500
Hi this is the structure and data of my table.
I need to create a mysql query for hotel booking, that would match in database user entered data:
Date when they want to checkin and checkout
Room type
For Ex:
If user selects Hotel with luxury room based on these dates (2012-08-30 to 2012-09-04)
the total cost would be (10000*2) for 30th and 31st Aug + (9500*3) for 1st,2nd and 3rd Sep(4th checkout day don't include)
that means total price will be 20000+28500=48500
So query should filter total price based on the Hotel_id,Room_id,Start_date,End_date and Price
Thanks

Use this solution:
SELECT SUM(
CASE WHEN a.Start_date = b.min_sd AND a.Start_date <> b.max_sd THEN
(DATEDIFF(a.End_date, '2012-08-30')+1) * a.Price
WHEN a.Start_date = b.max_sd AND a.Start_date <> b.min_sd THEN
DATEDIFF('2012-09-04', a.Start_date) * a.Price
WHEN (a.Start_date,a.Start_date) IN ((b.min_sd,b.max_sd)) THEN
(DATEDIFF('2012-09-04', '2012-08-30')+1) * a.Price
WHEN a.Start_date NOT IN (b.min_sd, b.max_sd) THEN
(DATEDIFF(a.End_date, a.Start_date)+1) * a.Price
END
) AS totalprice
FROM rooms a
CROSS JOIN (
SELECT MIN(Start_date) AS min_sd,
MAX(Start_date) AS max_sd
FROM rooms
WHERE Room_type = 'luxury' AND
End_date >= '2012-08-30' AND
Start_date <= '2012-09-04'
) b
WHERE a.Room_type = 'luxury' AND
a.End_date >= '2012-08-30' AND
a.Start_date <= '2012-09-04'
Replace occurances of 2012-08-30 and 2012-09-04 with your input start and end dates respectively.
This will account for start and end dates being in the same month as well as spanning across multiple months.
SQLFiddle Demo

You can use MySQL's BETWEEN ... AND ...
operator to find the date ranges in which the desired booking falls (remember to take one day off of the given checkout
date as, like you say, there is no night's stay), then group the results by room and take the
SUM() of price times number of nights (which can
be calculated using MySQL's LEAST() and
GREATEST() functions):
SELECT Room_id,
SUM(Price * (1 + DATEDIFF(
LEAST(End_date, '2012-09-04' - INTERVAL 1 DAY),
GREATEST(Start_date, '2012-08-30')
))) AS Total
FROM mytable
WHERE Room_type = 'luxury' AND (
'2012-09-04' - INTERVAL 1 DAY
BETWEEN Start_date AND End_date
OR '2012-08-30' BETWEEN Start_date AND End_date
)
GROUP BY Room_id
See it on sqlfidde.

try this:
set #Hotel_id :=13;
set #Room_id :=3;
set #Start_date :='2012-08-30' ;
set #End_date :='2012-09-04';
select sum(b.TotalPrice-b.deductions) as total_cost from
( select a.Price,a.StartDate,a.EndDate,price*(DATEDIFF(a.EndDate,a.StartDate)+1) as TotalPrice
,case when a.EndDate=#End_date then a.Price else 0 end as deductions
from
(select price,case when #Start_date>=Start_date then #Start_date else Start_date end as StartDate
,case when #End_date<=End_date then #End_date else End_date end as EndDate
from h_booking h1
where Hotel_id=#Hotel_id
and Room_id=#Room_id
and (#Start_date between Start_date and End_date or #End_date between Start_date and End_date ))a )b

Related

Get the minimum amount in a week

Hi I have a property_rate table and values like
id start_date end_date rate
1 2017-11-19 2017-11-21 100
2 2017-11-22 2017-11-24 300
3 2017-11-25 2017-11-28 500
4 2017-11-29 2017-11-30 200
Here i want to get the row having lowest price in a week like
id start_date end_date rate
1 2017-11-19 2017-11-21 100
What i have tried the below query but it is returning null
SELECT MIN(rate_per_week) AS min_value FROM property_rate WHERE start_date >= UNIX_TIMESTAMP(CURDATE() - INTERVAL 7 DAY)
Try this:
SELECT id, start_date, end_date, rate FROM <Table> ORDER BY rate ASC LIMIT <amount of rows to show>;
You have to place your table name.
here is the working code
SELECT * FROM `table name` ORDER BY `rate` ASC LIMIT 1
I have achieved it by below query
SELECT MIN(rate_per_week) AS min_value FROM property_rate WHERE start_date >= '2017-21-11' group by week('2017-21-11')

MySQL + PHP: Display items per week?

This has me a stummped...
If I have a this MySQL table:
UserId | Commission | Date Of Commission
1 | 200.00 | 2014-02-12
1 | 50.00 | 2014-04-01
2 | 10.00 | 2014-04-05
and I would like to display the Total Commission for a specific user per week starting from his/her first record, and display 0 for that range if there's no record.
how would I go about it?
Sample Output
UserId | Date Range | Total Commission
1 | 02/10/14 - 02/16/14 | 200.00
1 | 02/17/14 - 02/23/14 | 0.00
...
1 | 03/31/14 - 04/06/14 | 50.00
I'm not a seasoned coder so any help will be much appreciated.
Thanks!
Edit:
I have tried this:
SELECT IFNULL(SUM(Commisssion),0) Total ,DATE_SUB(`DateOfCommission`,INTERVAL 7 DAY)
AS RangStart,DATE_SUB(`DateOfCommission`,INTERVAL 1 DAY) AS RangeEnd
FROM `comms` WHERE `UserId` = '$UserID' GROUP BY DATE(`DateOfCommission`) DESC
but it starts the week with whatever date the first record was entered..
This is very tricky to accomplish. Here is what I managed to do with small modifications it should work they way it needs to be. I have done it for userid = 1 and this could be done for other users as well.
In the query I have 2 lines
where a.Date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()
and
WHERE date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()
The query will try to generate the list of dates using the min() date of transaction for the user till today. Instead of now() this could be used as max() date of transaction for the user as well.
select
t1.date_range,
coalesce(SUM(t1.Commission+t2.Commission), 0) AS Commission
from
(
select
a.Date as date,
concat(
DATE_ADD(a.Date, INTERVAL(1-DAYOFWEEK(a.Date)) +1 DAY),
' - ',
DATE_ADD(a.Date, INTERVAL(7- DAYOFWEEK(a.Date)) +1 DAY)
) as date_range,
'0' as Commission
from (
select curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY as date
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.Date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()
)t1
left join
(
SELECT date ,
coalesce(SUM(Commission), 0) AS Commission
FROM transactions
WHERE date BETWEEN (select min(date) from transactions where UserId = 1) AND NOW()
AND UserId = 1
GROUP BY date
)t2
on t2.date = t1.date
group by t1.date_range
order by t1.date_range asc
DEMO
So, this is sort of an algorithm you could use:
$Result = select distinct userid from table(this will fetch all userids from table)
while(There are rows in $Result)
{
$Userid = $Result['userid']
$StartDateRes = mysql_query(select Date, WEEKOFYEAR(Date) as week from table where userid = Userid order by date asc limit 1)
$StartDateRow = mysql_fetch_assoc($StartDateRes)
$StartDate = $StartDateRes['Date']
$StartWeekNumber = $StartDateRes['week']
$EndDateRes = mysql_query(select Date, WEEKOFYEAR(Date) as week from table where userid = Userid order by date desc limit 1)
$EndDateRow = mysql_fetch_assoc($EndDateRes)
$EndDate = $EndDateRes['Date']
$EndWeekNumber = $EndWeekRes['week']
for($i=$StartWeekNumber; $i<=$EndWeekNumber; $i++)
{
$StartDateOfWeek = FuncToFindStartDateOfWeek($i)
$EndDateOfWeek = FuncToFindEndDateOfWeek($i)
$Result2 = mysql_query(select sum(commission) as sum from table where date between StartDateOfWeek and EndDateOfWeek group by userid)
$Row2= mysql_fetch_assoc($Result2)
$Sum = $Row2['sum']
mysql_query("insert into OutputTable values($UserId, $StartDateOfWeek. '-'. $EndDateOfWeek ,$Sum");
}
}
SELECT UserId, COALESCE(SUM(Commission),0), YEARWEEK(DateOfCommission) AS TheWeek
GROUP BY UserId, TheWeek
ORDER BY UserId, TheWeek;
This will not print the nice date range, but should get you started in a SQL-only direction where the sum is broken down by the week of the year. I think you could take it from this point to add the nicer formatting of the Year/Week column. YEARWEEK() should give you pretty fast results.
The easiest way I can think of doing this as follows
Step 1: Get the date of the first record
"SELECT dateofcommission FROM comissionstable WHERE id='userid' ORDER BY dateofcommission ASC LIMIT 1"
The above query will return the first date of commission only
Step 2: Create a loop which starts from the date you got in Step 1 and continue the loop till the date is greater than or equal to today's date. Increment this date using PHP date function.
date('Y-m-d', strtotime($dateofcommission. ' + 7 days'));
Step 3: In this loop you can get the commission with-in the starting date and ending date. Starting date will be the date before adding 7 days and ending date will be the one after you have added 7 days.
SELECT SUM(commission) FROM commissiontable WHERE dateofcommission>= startingdate AND dateofcomission < endingdate AND id='userid'
The above logic should work. If you end up having some issues with this logic feel free to post in comments. I would be happy to help
The following is another solution
function getStartAndEndDate($week, $year) {
$time = strtotime("1 January $year", time());
$day = date('w', $time);
$time += ((7*$week)+1-$day)*24*3600;
$return[0] = date('Y-n-j', $time);
$time += 6*24*3600;
$return[1] = date('Y-n-j', $time);
return $return;
}
$query = mysqli_query($con, "SELECT userid, COALESCE( SUM( commission ) , 0 ) AS thecommission , YEARWEEK( doc ) AS TheWeek FROM commission GROUP BY userid, TheWeek ORDER BY userid, TheWeek");
while ($array = mysqli_fetch_array($query)) {
$test = $array['TheWeek'];
$store_array = getStartAndEndDate(substr($test,4,2), substr($test,0,4));
echo $array['userid']."-".$array['thecommission']."-".$store_array[0]."/".$store_array[1]."<br>";
}

MYSQL select query with multiple tables

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

MySQL BETWEEN DATE RANGE

I have a scenario where I need to pull up delivery dates based on a table below (Example)
job_id | delivery_date
1 | 2013-01-12
2 | 2013-01-25
3 | 2013-02-15
What I'm trying to do is show the user all the delivery dates that start with the earliest (in this case it would be 2013-01-12) and add an another 21 days to that. Basically, the output I would expect it to show of course, the earliest date being the starting date 2013-01-12 and 2013-01-25. The dates past the February date are of no importance since they're not in my 21 date range. If it were a 5 day range, for example, then of course 2013-01-25 would not be included and only the earliest date would appear.
Here is main SQL clause I have which only shows jobs starting this year forward:
SELECT date, delivery_date
FROM `job_sheet`
WHERE print_status IS NULL
AND job_sheet.date>'2013-01-01'
Is it possible to accomplish this with 1 SQL query, or must I go with a mix of PHP as well?
You can use the following:
select *
from job_sheet
where print_status IS NULL
and delivery_date >= (select min(delivery_date)
from job_sheet)
and delivery_date <= (select date_add(min(delivery_date), interval 21 day)
from job_sheet)
See SQL Fiddle with Demo
If you are worried about the dates not being correct, if you use a query then it might be best to pass in the start date to your query, then add 21 days to get the end date. Similar to this:
set #a='2013-01-01';
select *
from job_sheet
where delivery_date >= #a
and delivery_date <= date_add(#a, interval 21 day)
See SQL Fiddle with Demo
SELECT date,
delivery_date
FROM job_sheet
WHERE print_status IS NULL
AND job_sheet.date BETWEEN (SELECT MIN(date) FROM job_sheet) AND
(SELECT MIN(date) FROM job_sheet) + INTERVAL 21 DAY
SELECT j.job_id
, j.delivery_date
FROM `job_sheet` j
JOIN ( SELECT MIN(d.delivery_date) AS earliest_date
FROM `job_sheet` d
WHERE d.delivery_date >= '2013-01-01'
) e
ON j.delivery_date >= e.earliest_date
AND j.delivery_date < DATE_ADD(e.earliest_date, INTERVAL 22 DAY)
AND j.print_status IS NULL
ORDER BY j.delivery_date
(The original query has a predicate on job_sheet.date; the query above references the d.delivery_date... change that if it is supposed to be referencing the date column instaed.)
If the intent is to only show delivery_date values from today forward, then change the literal '2013-01-01' to an expression that returns the current date, e.g. DATE(NOW())

Find first free date in agenda

in my component I have created an agenda, where user can save their appointments.
Agenda table is quite simple: there is a title, description and start/end datetime fields.
When a user adds a new event, I'd wish to hint him with the first empty spot.
How can I achieve that?
Is that possible with a single/bunch queries, or I have to create a loop until I find the first empty spot?
For example this is my table:
| ID | Start date | End date |
| 1 | 2012-06-14 09:00:00 | 2012-06-14 09:32:00 |
| 2 | 2012-06-14 15:00:00 | 2012-06-14 15:45:00 |
| 3 | 2012-06-14 18:20:00 | 2012-06-14 18:55:00 |
The first free datetime should be 2012-06-14 09:33:00, how can I fetch this date?
Interesting challenge to be done in one query :) Took me a while but I came with solution. With these assumptions:
minimum appointment duration is 30 minutes
appointment starts and ends in one day
day starts at 9:00 and ends at 17:00
minimum interval between ending and starting times is 1 minute
There are 4 cases to take into consideration:
There is some free time in the morning
There is some free time during the day
First free slot is next day to the last appointment in db
You have all free slots from now on
So yo have to select minimum form these dates.
And the query:
SELECT
MIN(next_start_date) AS next_start_date
FROM
(
(
SELECT
DATE_FORMAT(t1.start_date,'%Y-%m-%d 09:00:00') AS next_start_date
FROM
agenda t1
WHERE
t1.start_date > NOW()
AND
TIME(t1.start_date) > '09:30:00'
AND
NOT EXISTS(
SELECT 1 FROM agenda t2 WHERE DATE(t2.start_date) = DATE(t1.start_date) AND TIME(t2.start_date) <= '09:30:00'
)
LIMIT 1
)
UNION
(
SELECT
t3.end_date + INTERVAL 1 MINUTE AS next_start_date
FROM
agenda t3
WHERE
t3.start_date > NOW()
AND
TIME(t3.start_date) >= '09:00:00'
AND
TIME(t3.end_date) < '16:30:00'
AND NOT EXISTS
(SELECT 1 FROM agenda t4 WHERE TIMESTAMPDIFF(MINUTE,t3.end_date,t4.start_date) BETWEEN 0 AND 30 )
ORDER BY
t3.start_date ASC
LIMIT 1
)
UNION
(
SELECT CONCAT(CAST(DATE((SELECT MAX(t5.start_date) + INTERVAL 1 DAY FROM agenda t5 WHERE t5.start_date > NOW())) AS CHAR), ' 09:00:00') AS next_start_date
)
UNION
(
SELECT
IF(
TIME(NOW()) < '09:00:00',
DATE_FORMAT(NOW(),'%Y-%m-%d 09:00:00'),
IF(
TIME(NOW()) < '16:30',
NOW(),
DATE_FORMAT(NOW() + INTERVAL 1 DAY,'%Y-%m-%d 09:00:00' )
)
) AS next_start_date
FROM
(SELECT 1) t6
WHERE NOT EXISTS(
SELECT 1 FROM agenda t7 WHERE t7.start_date > NOW()
)
LIMIT 1
)
) t
Of course it is not perfect - when there next free slot occurs in the next day, there's a chance that it is Saturday or other day off from work. Taking it into consideration, the best way will be to check if returned is valid work day, if not then repeat the query with NOW() replaced by datestring of next valid work day.

Categories