Question

Shalini, who works as a database designer in the hotel industry, has created a table named Guest to keep track of guest details as shown below:

Table : Guest

GuestID	GuestName	RoomNumber	CheckInDate	Charges
G101	Harish	101	2025-04-03	3000
G102	Sunita	101	2025-04-03	3000
G103	Ramesh	102	2025-05-04	5000
G104	Bhumika	103	2025-06-02	3500

Write a suitable SQL query for the following:

I. Display last 3 characters of guest name in upper case.

II. Display the name of the guest along with the day name of check-in date.

III. Display the remainder when charges are divided by 1000.

IV. Extract and display three characters, starting from the second character, of each guest name.

Hint:

Use RIGHT with UPPER, DAYNAME for the weekday, MOD for the remainder and MID or SUBSTR starting at position 2 with length 3.

Answer 1

Step 1: Read the table:
The table Guest has 4 rows. GuestName holds text, CheckInDate holds a date and Charges holds a number.
Each part needs one single-row function applied to a column.

Step 2: Part (I) - last 3 characters in upper case:
RIGHT(string, n) gives the last n characters. UPPER() changes them to capital letters. We put RIGHT inside UPPER.

SELECT UPPER(RIGHT(GuestName, 3)) FROM Guest;

Output: ISH, ITA, ESH, IKA.

Step 3: Part (II) - name with the day name of check-in date:
DAYNAME(date) gives the weekday as a word.

SELECT GuestName, DAYNAME(CheckInDate) FROM Guest;

Output: Harish - Thursday, Sunita - Thursday, Ramesh - Sunday, Bhumika - Monday.

Step 4: Part (III) - remainder of charges divided by 1000:
MOD(a, b) gives the remainder of a divided by b.

SELECT MOD(Charges, 1000) FROM Guest;

3000 gives 0, 3000 gives 0, 5000 gives 0 and 3500 gives 500.

Step 5: Part (IV) - three characters from the second character:
MID(string, start, length) or SUBSTR(string, start, length) starts counting at position 1. We start at 2 and take 3 characters.

SELECT MID(GuestName, 2, 3) FROM Guest;

Harish gives ari, Sunita gives uni, Ramesh gives ame and Bhumika gives hum.

Final Answer:
I. SELECT UPPER(RIGHT(GuestName, 3)) FROM Guest;
II. SELECT GuestName, DAYNAME(CheckInDate) FROM Guest;
III. SELECT MOD(Charges, 1000) FROM Guest;
IV. SELECT MID(GuestName, 2, 3) FROM Guest;