1. SQL queries ordering and grouping and joins

2. SQL query - ordering
In a query producing a large result, it may be beneficial to order the result
SQL allows us to order the result by any of the fields in the result
We use the keyword ORDER BY
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3. SQL query - ordering SELECT <fieldlist> FROM <tablename>
Which fields do I want
SELECT <fieldlist>
FROM <tablename>
WHERE <condition>
ORDER BY <fieldname>
From what table do I want the fields
What conditions must the fields fulfill
What order are the results sorted in
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4. HotelDB HOTEL: (Hotel_No, Name, Address)
ROOM: (Room_No, Hotel_No, Types, Price)
BOOKING: (Hotel_No, Guest_No, Date_From, Date_To, Room_No)
GUEST: (Guest_No, Name, Address)
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5. SQL query - ordering SELECT * FROM Hotel ORDER BY Name;
SELECT Guest_No, Date_From, Room_no FROM booking WHERE Hotel_no = 1
ORDER BY Date_From;
SELECT * FROM guest ORDER BY Address;
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6. SQL query - ordering
We can even specifiy more than one field for ordering – secondary fields used if primary fields are identical
We can choose between descending and ascending order, using the keywords DESC and ASC, respectively
SELECT Guest_No, Date_From, Room_no
FROM booking WHERE Hotel_no = 1
ORDER BY Guest_No DESC, Date_From ASC;
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7. SQL query - functions
We can even do some (simple) arithmetic in SQL, using a basic set of functions
COUNT
SUM
AVG
MIN
MAX
These are called aggregate functions
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8. SQL query - functions
A aggregate function works on the values of a specific field (column)
The set of values is determined by the search conditions
How many rooms have Hotel no 1
SELECT count(*) FROM room WHERE Hotel_No = 1;
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9. SQL query - functions This query can also be written as
SELECT count(*) AS Number_of_Rooms
FROM room WHERE Hotel_No = 1;
The AS keyword allows us to rename a column in the search result
Only cosmetic, but useful…
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10. SQL query - functions SELECT
COUNT(*) AS Number_of_Rooms, AVG(Price) as Avg_price,
Max(Price)
FROM room
where Hotel_no = 1;
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11. Exercise 4 – SQL queries Use the Hotel_DB database.
With the data in place, run the below queries on the database
SELECT * FROM Booking ORDER BY Hotel_no ASC, Room_NO DESC
SELECT Name, Address FROM Hotel ORDER BY Hotel_No;
SELECT MAX(Price) AS maxPrice FROM room WHERE Types = 'D';
Now formulate queries yourself, in order to retrieve the below data:
Get all bookings from hotel_no 2 (oldest first)
Get a sorted list of name and address of Guests
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12. SQL query - grouping
The aggregate functions are good for calculating properties for the entire result of the search
We may sometimes wish to find proper-ties for a certain group within the result
This can be done using WHERE…
…but can be cumbersome if the groups are very numerous
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13. SQL query - grouping Syntax: SELECT <fieldlist>
FROM <tablename>
GROUP BY <fieldlist>
Produces a result for each group, specified in the field list
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14. SQL query - grouping
”Find the total number of booking made for each hotel”
SELECT count(*)as bookings , hotel_no
from booking group by hotel_no
order by hotel_no;
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15. SQL query - grouping
In general; only include groups which fulfill some criteria
This can be done using the HAVING keyword
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16. SQL query - grouping Syntax: SELECT <fieldlist>
FROM <tablename>
GROUP BY <fieldlist>
HAVING <criteria>
SELECT count(*), types, sum(price)
FROM room
GROUP BY types
HAVING sum(price)>3000 ;
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17. SQL query - grouping But wait… …isn’t HAVING the same as WHERE..?
Not quite
WHERE is for filtering out specific records
HAVING is for filtering out specific groups from the final result
We cannot use an aggregate function in a WHERE clause
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18. SQL query – beyond one table
So far, we have only applied queries to a single table
It is possible – and very useful – to make queries applied to several tables
Can answer more complex questions…
…but queries also tend to become more complex themselves 
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19. SQL query – beyond one table
We use a HotelDB database as example
Room
Guest
Booking
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20. SQL query – subqueries
Suppose we wish to answer a (complex) question like this:
Find room_no, room type, price for all hotels in Roskilde
This question cannot be answered by a single query (as we know it…)
Information is spread out in multiple tables
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21. SQL query – subqueries SELECT room_no, types, price FROM room
WHERE hotel_no IN
(SELECT hotel_no
FROM hotel
WHERE address LIKE '%Roskilde%');
Use the IN keyword when more than one result from the subquery
2011
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22. SQL query – subqueries
Result from inner query (the subquery) is used as input to outer query
The inner query produces a result table – just as any other query – which is then used by the outer query
Very convenient – but some complications
Name clash
Multiple results
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23. SQL query – subqueries
When we query more than one table, field names from different tables might be identical – a name clash
We can qualify a field name by prefixing it with the table name
Room.hotel_no
Hotel.hotel_no
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24. SQL query – subqueries SELECT room_no, types, price, Hotel.name
FROM room, hotel
WHERE Room.hotel_no IN
(SELECT hotel_no
FROM hotel
WHERE address LIKE '%Roskilde%');
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25. SQL query – joining
Another approach to multi-table queries is to use join
To ”join” tables is to make a kind of ”multiplication” of the tables
Simple case: Join without any conditions:
SELECT *
FROM Room, Hotel
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26. SQL query – joining
If the previous query is run against the HotelDB database
How many rows
All combinations of Hotel records and Room records
All fields from both tables are included
This is rarely what we want…
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27. SQL query – joining
In almost all cases, we wish to pick out those rows where certain fields match
Example: Find the room_no, type and price of the room, and the name of the Hotel, for each Hotel from Roskilde
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28. SQL query – joining SELECT Room_No, types, name, price
FROM room r, hotel h
WHERE h.hotel_no = r.hotel_no
AND address LIKE '%Roskilde%';
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29. SQL query – joining
This is a quite common ”pattern” for multi-table queries
Tables representing relations only contain identifiers (keys in other tables)
”Real data” is contained in tables representing entities (identifiers are keys)
Obtaining ”real data” for relations requires use of join, with matching of key fields
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30. Different SQL JOINs INNER JOIN:
Returns all rows when there is at least one match in BOTH tables
LEFT JOIN:
Return all rows from the left table, and the matched rows from the right table
RIGHT JOIN:
Return all rows from the right table, and the matched rows from the left table
FULL JOIN:
Return all rows when there is a match in ONE of the tables
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31. Inner Join
Selects all rows from both tables as long as there is a match between the columns in both tables.
Syntax:
SELECT column_name(s) FROM table1 INNER JOIN table2 ON table1.column_name=table2.column_name;
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32. Inner Join SELECT Room_No, types, name, price FROM room
INNER JOIN hotel
ON room.hotel_no =hotel.hotel_no
Returns all rows when there is matching Hotel_No in BOTH tables
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33. LEFT JOIN
Return all rows from the left table, and the matched rows from the right table
Syntax:
SELECT column_name(s) FROM table1 LEFT JOIN table2 ON table1.column_name=table2.column_name;
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34. LEFT JOIN List all Guests and their Bookings if they have any:
SELECT Guest.Guest_No, Guest.Name, Booking.Date_From, Booking.Date_To
FROM Guest
LEFT JOIN Booking
ON Guest.Guest_No = Booking.Guest_No
ORDER BY Guest.Guest_No;
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35. RIGHT JOIN
Returns all rows from the right table (table2), with the matching rows in the left tab
Syntax:
SELECT column_name(s) FROM table1 RIGHT JOIN table2 ON table1.column_name=table2.column_namele (table1).
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36. RIGHT JOIN List all Hotel and the Rooms in the Hotels
SELECT Room.Room_No, Hotel.Hotel_No, Hotel.Name
FROM Room
RIGHT JOIN Hotel
ON Room.Hotel_No=Hotel.Hotel_No
ORDER BY Hotel.Hotel_No;
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37. FULL OUTER JOIN
Returns all rows from the left table (table1) and from the right table (table2).
The FULL OUTER JOIN keyword combines the result of both LEFT and RIGHT joins.
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38. FULL OUTER JOIN Syntax:
SELECT column_name(s) FROM table1 FULL OUTER JOIN table2 ON table1.column_name=table2.column_name;
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39. FULL OUTER JOIN
SELECT Guest.Guest_No, Guest.Name, Booking.Date_From, Booking.Date_To
FROM Guest
FULL OUTER JOIN Booking
ON Guest.Guest_No = Booking.Guest_No
ORDER BY Guest.Guest_No;
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