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Hi, This is kind of last minute, I have a day and a half left to figure
this out. I'm working on a project using mssqlserver. We are
creating a ticket sales system, as part of the system, I need to be
able to do a search for specific tickets withing price ranges,
different locations within the theaters, etc. etc.
My problem is in the search one of the criteria is to search for a
group of seats together. For example let's say someone would like to
purchase four tickets to an event they are obviously going to want all
four tickets together in the same row/area...
I am trying to figure out some way in which, if in the search criteria
say "4" was entered for the number of seats they want grouped together,
I can return only the different sections/rows that have "4" or more
seats grouped together that have an "available" status. Any help would
be greatly appreciated.
Thankyou
Spidey  
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I thought I'd include the two main tables I feel I'm dealing with ...
If you feel I need something a little more than what I've got here,
just let me know..
Table 1: '''EventSeat''' (This is the information for each individual
seat for each event. An event would be for example the 7:00 showing of
a theatrical play.)
'eventID'
'seatNumber'
'status'
'rowNumber'
'sectionID'
Table 2: '''PhysicalRow'''
'rowNumber'
'sectionID'
'catID' (Price Category ID)
'numberSeats' (Number of Seats in Row)
Thanks ahead of time, for any help you can offer.
~~~~ Spidey ~~~~  
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AFAIK, there is no simple way using pure SQL.
You have to write stored procedure, which scans all free seats. Then it
checks all 3 (in your case) neighbors for each such seat.
It's ugly, time consuming (O(free_seats_count ^ reserved_places)), but
imho the only way. You can optimize it by creating column
largest_free_group in Table 2 and update this column in a trigger.
Hope it's clear enough :)
tv
Orion napsal(a): I thought I'd include the two main tables I feel I'm dealing with ... If you feel I need something a little more than what I've got here, just let me know..
Table 1: '''EventSeat''' (This is the information for each individual seat for each event. An event would be for example the 7:00 showing of a theatrical play.) 'eventID' 'seatNumber' 'status' 'rowNumber' 'sectionID'
Table 2: '''PhysicalRow''' 'rowNumber' 'sectionID' 'catID' (Price Category ID) 'numberSeats' (Number of Seats in Row)
Thanks ahead of time, for any help you can offer. ~~~~ Spidey ~~~~  
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Here's an example that might help you:
CREATE TABLE BookedSeats (event_dt DATETIME NOT NULL, seat_no INTEGER
NOT NULL, booking_no INTEGER NOT NULL /* REFERENCES Bookings
(booking_no) */, PRIMARY KEY (event_dt, seat_no))
INSERT INTO BookedSeats VALUES ('20051231',1,1111)
INSERT INTO BookedSeats VALUES ('20051231',2,2222)
INSERT INTO BookedSeats VALUES ('20051231',6,3333)
SELECT T1.seat_no, COALESCE(MIN(T2.seat_no),999)
FROM BookedSeats AS T1
LEFT JOIN BookedSeats AS T2
ON T1.seat_no < T2.seat_no
AND T1.event_dt = T2.event_dt
GROUP BY T1.seat_no
HAVING COALESCE(MIN(T2.seat_no),999)T1.seat_no = 4 /* no. of required seats */

David Portas
SQL Server MVP
  
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This was an example in SQL FOR SMARTIES about ten years ago.
24.01. Finding Subregions of Size (n)
This example is adapted from SQL and Its Applications (Lorie and
Daudenarde 1991). You are given a table of theater seats, defined by
CREATE TABLE Theater
(seat_nbr INTEGER NOT NULL PRIMARY KEY,  sequencing number
occupancy_status CHAR(1) NOT NULL  values
CONSTRAINT valid_occupancy_status
CHECK (occupancy_status IN ('A', 'S'));
where an occupancy_status code of 'A' means available and 'S' means
sold. Your problem is to write a query that will return the subregions
of (n) consecutive seats still available. Assume that consecutive
seat_nbrs means that the seats are also consecutive for a moment,
ignoring rows of seating where seat_nbr(n) and seat_nbr((n) + 1) might
be on different physical theater rows. For (n) = 3, we can write a
selfJOIN query, thus:
SELECT T1.seat_nbr, T2.seat_nbr, T3.seat_nbr
FROM Theater AT T1, Theater AT T2, Theater AT T3
WHERE T1.occupancy_status = 'A'
AND T2.occupancy_status = 'A'
AND T3.occupancy_status = 'A'
AND T2.seat_nbr = T1.seat_nbr + 1
AND T3.seat_nbr = T2.seat_nbr + 1;
The trouble with this answer is that it works only for (n = 3) and for
nothing else. This pattern can be extended for any (n), but what we
really want is a generalized query where we can use (n) as a parameter
to the query.
The solution given by Lorie and Daudenarde starts with a given seat_nbr
and looks at all the available seats between it and ((n)  1) seats
further up. The real trick is switching from the Englishlanguage
statement "All seats between here and there are available" to the
passivevoice version, "Available is the occupancy_status of all the
seats between here and there", so that you can see the query.
SELECT seat_nbr, ' thru ', (seat_nbr + (:(n)  1))
FROM Theater AS T1
WHERE occupancy_status = 'A'
AND 'A' = ALL (SELECT occupancy_status
FROM Theater AS T2
WHERE T2.seat_nbr > T1.seat_nbr
AND T2.seat_nbr <= T1.seat_nbr + (:(n)  1));
Please notice that this returns subregions. That is, if seats
(1, 2, 3, 4, 5) are available, this query will return (1, 2, 3),
(2, 3, 4), and (3, 4, 5) as its result set.
24.02. Numbering Regions
Instead of looking for a region, we want to number the regions in the
order in which they appear. For example, given a view or table with a
payment history we want to break it into grouping of behavior, say
whether or not the payments were on time or late.
CREATE TABLE PaymentHistory
(payment_nbr INTEGER NOT NULL PRIMARY KEY,
paid_on_time CHAR(1) DEFAULT 'Y' NOT NULL
CHECK(paid_on_time IN ('Y', 'N')));
INSERT INTO PaymentHistory
VALUES (1006, 'Y'), (1005, 'Y'),
(1004, 'N'),
(1003, 'Y'), (1002, 'Y'), (1001, 'Y'),
(1000, 'N');
The results we want assign a grouping number to each run of
ontime/late payments, thus.
Results
grping payment_nbr paid_on_time
===============================
1 1006 'Y'
1 1005 'Y'
2 1004 'N'
3 1003 'Y'
3 1002 'Y'
3 1001 'Y'
4 1000 'N'
A solution by Hugo Kornelis depends on the payments always being
numbered consecutively.
SELECT (SELECT COUNT(*)
FROM PaymentHistory AS H2,
PaymentHistory AS H3
WHERE H3.payment_nbr = H2.payment_nbr + 1
AND H3.paid_on_time <> H2.paid_on_time
AND H2.payment_nbr >= H1.payment_nbr) + 1 AS grping,
payment_nbr, paid_on_time
FROM PaymentHistory AS H1;
This can be modified for more types of behavior.
24.03. Finding Regions of Maximum Size
A query to find a region, rather than a subregion of a known size, of
seats was presented in SQL Forum (Rozenshtein, Abramovich, and Birger
1993).
SELECT T1.seat_nbr, ' thru ', T2.seat_nbr
FROM Theater AS T1, Theater AS T2
WHERE T1.seat_nbr < T2.seat_nbr
AND NOT EXISTS
(SELECT *
FROM Theater AS T3
WHERE (T3.seat_nbr BETWEEN T1.seat_nbr AND T2.seat_nbr
AND T3.occupancy_status <> 'A')
OR (T3.seat_nbr = T2.seat_nbr + 1
AND T3.occupancy_status = 'A')
OR (T3.seat_nbr = T1.seat_nbr  1
AND T3.occupancy_status = 'A'));
The trick here is to look for the starting and ending seats in the
region. The starting seat_nbr of a region is to the right of a sold
seat_nbr and the ending seat_nbr is to the left of a sold seat_nbr. No
seat_nbr between the start and the end has been sold.
If you only keep the available seat_nbr's in a table, the solution is a
bit easier. It is also a more general problem that applies to any
table of sequential, possibly noncontiguous, data:
CREATE TABLE AvailableSeating
(seat_nbr INTEGER NOT NULL
CONSTRAINT valid_seat_nbr
CHECK (seat_nbr BETWEEN 001 AND 999));
INSERT INTO Seatings
VALUES (199), (200), (201), (202), (204),
(210), (211), (212), (214), (218);
where you need to create a result which will show the start and finish
values of each sequence in the table, thus:
Results
start finish
============
199 202
204 204
210 212
214 214
218 218
This is a common way of finding the missing values in a sequence of
tickets sold, unaccounted for invoices and so forth. Imagine a number
line with closed dots for the numbers that are in the table and open
dots for the numbers that are not. What you see about a sequence?
Well, we can start with a fact that anyone who has done inventory
knows. The number of elements in a sequence is equal to the ending
sequence number minus the starting sequence number plus one. This is a
basic property of ordinal numbers:
(finish  start + 1) = Length of open seats
This tells us that we need to have a selfJOIN with two copies of the
table, one for the starting value and one for the ending value of each
sequence. Once we have those two items, we can compute the length with
our formula and see if it is equal to the count of the items between
the start and finish.
SELECT S1.seat_nbr, MAX(S2.seat_nbr)  start and rightmost item
FROM AvailableSeating AS S1
INNER JOIN
AvailableSeating AS S2  selfjoin
ON S1.seat_nbr <= S2.seat_nbr
AND (S2.seat_nbr  S1.seat_nbr + 1)  formula for length
= (SELECT COUNT(*)  items in the sequence
FROM AvailableSeating AS S3
WHERE S3.seat_nbr BETWEEN S1.seat_nbr AND S2.seat_nbr)
AND NOT EXISTS (SELECT *
FROM AvailableSeating AS S4
WHERE S1.seat_nbr  1 =
S4.seat_nbr)
GROUP BY S1.seat_nbr;
Finally, we need to be sure that we have the furthest item to the right
as the end item. Each sequence of (n) items has (n) subsequences that
all start with the same item. So we finally do a GROUP BY on the
starting item and use a MAX() to get the rightmost value.
However, there is a faster version with three tables. This solution is
based on another property of the longest possible sequences. If you
look to the right of the last item, you do not find anything.
Likewise, if you look to the left of the first item, you do not find
anything either. These missing items that are "just over the border"
define a sequence by framing it. There also cannot be any "gaps" 
missing items  inside those borders. That translates into SQL as:
SELECT S1.seat_nbr, MIN(S2.seat_nbr) start and leftmost border
FROM AvailableSeating AS S1, AvailableSeating AS S2
WHERE S1.seat_nbr <= S2.seat_nbr
AND NOT EXISTS  border items of the sequence
(SELECT *
FROM AvailableSeating AS S3
WHERE S3.seat_nbr NOT BETWEEN S1.seat_nbr AND S2.seat_nbr
AND (S3.seat_nbr = S1.seat_nbr  1
OR S3.seat_nbr = S2.seat_nbr + 1))
GROUP BY S1.seat_nbr;
We do not have to worry about getting the rightmost item in the
sequence, but we do have to worry about getting the leftmost border.
Once we do a GROUP BY, but use a MIN() to get what we want.
Since the second approach uses only three copies of the original table,
it should be a bit faster. Also, the EXISTS() predicates can often
take advantage of indexing and thus run faster than subquery
expressions which require a table scan.
Michel Walsh came up with two novel ways of getting the range of seat
numbers that have been used in the table. He saw that the difference
between the value and its rank is a constant for all values in the same
consecutive sequence, so we just have to group, and count, on the value
minus its rank to get the various consecutive runs (or just keep the
maximum). It is so simple, an example will show every thing:
data = {1, 2, 5, 6, 7, 8, 9, 11, 12, 22}
data rank (data_rank) AS absent
================================
1 1 0
2 2 0
5 3 2
6 4 2
7 5 2
8 6 2
9 7 2
11 8 3
12 9 3
22 10 12
absent COUNT(*)
================
0 2
2 5
3 2
12 1
so, the maximum contiguous sequence is 5 (for rows having (data  rank)
= 2). Rank is defined as how many values are less or equal to the
actual value, with the assumption of a set of integers without repeated
values. This is the query.
SELECT X.absent, COUNT(*)
FROM (SELECT my_data,
(SELECT COUNT(*)
FROM Foobar AS F2
WHERE F2.my_data <= F1.my_data),
(SELECT COUNT(*)
FROM Foobar AS F2
WHERE F2.my_data <= F1.my_data)  F1.my_data
FROM Foobar AS F1)
AS X(my_data, rank, absent);
Playing with this basic idea, Mr. Walsh came up with this second query.
SELECT MIN(Z.seat_nbr), MAX(Z.seat_nbr)
FROM (SELECT S1.seat_nbr,
S1.seat_nbr
 (SELECT COUNT(*)
FROM Seating AS S2
WHERE S2.seat_nbr <= S1.seat_nbr)
FROM Seating AS S1)
AS Z (seat_nbr, dif_rank)
GROUP BY Z.dif_rank;
The derived table finds the lengths of the blocks of seats to the left
of each seat_nbr and uses that length to form groups.   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 2152
 replies: 4
 date asked: Jul 23 '05
