sppu-te-comp-content/DatabaseManagementSystems
sppu-te-comp-content/DatabaseManagementSystems
3
1
Fork 0
Code Issues Activity

Simplified queries for A2:

Browse Source 
- Creating tables on multiple lines for better understanding
- Changed names of columns to match names in assignment handout
- Updated some values so that each query in implementation gives valid output
- Lastly, simplified some queries and changed column names
This commit is contained in:
K 2024-11-01 02:22:03 +05:30
parent fa91c4f08f
commit 0c887415b3
Signed by: notkshitij
GPG Key ID: C5B8BC7530F8F43F
1 changed files with 100 additions and 71 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

171
Practical/Assignment-A2/Queries-A2.md Executable file → Normal file
View File

@ -13,74 +13,100 @@ USE Database_A2;
## Creating tables and specify primary keys:
```sql
CREATE TABLE Account(accountNum INT, branchName VARCHAR(50), balance INT, PRIMARY KEY (accountNum));
CREATE TABLE Branch(branchName VARCHAR(50), branchCity VARCHAR(50), assets INT, PRIMARY KEY (branchName));
CREATE TABLE Customer (customerName VARCHAR(50), customerStreet VARCHAR(50), customerCity VARCHAR(50), PRIMARY KEY (customerName));
CREATE TABLE Depositor (customerName VARCHAR(50), accountNum INT);
CREATE TABLE Loan (loanNum INT, branchName VARCHAR(50), amount INT, PRIMARY KEY (loanNum));
CREATE TABLE Borrower (customerName VARCHAR(50), loanNum INT);
CREATE TABLE Account (
acc_no INT,
branch_name VARCHAR(255),
balance INT,
PRIMARY KEY (acc_no)
);
CREATE TABLE Branch (
branch_name VARCHAR(255),
branch_city VARCHAR(255),
assets INT,
PRIMARY KEY (branch_name)
);
CREATE TABLE Customer (
cust_name VARCHAR(255),
cust_street VARCHAR(255),
cust_city VARCHAR(255),
PRIMARY KEY (cust_name)
);
CREATE TABLE Depositor (
cust_name VARCHAR(255),
acc_no INT
);
CREATE TABLE Loan (
loan_no INT,
branch_name VARCHAR(255),
amount INT,
PRIMARY KEY (loan_no)
);
CREATE TABLE Borrower (
cust_name VARCHAR(255),
loan_no INT
);
```
## Altering tables to specify foreign keys
```sql
ALTER TABLE Account ADD FOREIGN KEY (branchName) REFERENCES Branch(branchName);
ALTER TABLE Depositor ADD FOREIGN KEY (customerName) REFERENCES Customer (customerName);
ALTER TABLE Depositor ADD FOREIGN KEY (accountNum) REFERENCES Account (accountNum);
ALTER TABLE Loan ADD FOREIGN KEY (branchName) REFERENCES Branch (branchName);
ALTER TABLE Borrower ADD FOREIGN KEY (customerName) REFERENCES Customer (customerName);
ALTER TABLE Borrower ADD FOREIGN KEY (loanNum) REFERENCES Loan (loanNum);
ALTER TABLE Account ADD FOREIGN KEY (branch_name) REFERENCES Branch (branch_name);
ALTER TABLE Depositor ADD FOREIGN KEY (cust_name) REFERENCES Customer (cust_name);
ALTER TABLE Depositor ADD FOREIGN KEY (acc_no) REFERENCES Account (acc_no);
ALTER TABLE Loan ADD FOREIGN KEY (branch_name) REFERENCES Branch (branch_name);
ALTER TABLE Borrower ADD FOREIGN KEY (cust_name) REFERENCES Customer (cust_name);
ALTER TABLE Borrower ADD FOREIGN KEY (loan_no) REFERENCES Loan (loan_no);
```
## Adding data
```sql
INSERT INTO Branch (branchName, branchCity, assets) VALUES
("Dhole Patil", "Kharadi", 50000),
("Nagarwala", "Akurdi", 20000),
("Peachtree", "Wakad", 35000),
("Bishops", "Nigdi", 10000),
("Amanora", "Hadapsar", 60000);
INSERT INTO Branch (branch_name, branch_city, assets) VALUES
('Akurdi', 'Pune', 5000000),
('Nigdi', 'Pune', 3000000),
('Kharadi', 'Pune', 4000000),
('Hadapsar', 'Pune', 3500000),
('Viman Nagar', 'Pune', 4500000);
INSERT INTO Customer (customerName, customerStreet, customerCity) VALUES
("Kalas", "Airport Road", "Pune"),
("Mehul", "Shahdha", "Nandurbar"),
("Tanmay", "Porwal Road", "Pune"),
("Kshitij", "Hadapasar", "Pune"),
("Aditya", "Mira RD", "Mumbai"),
("Himanshu", "Smart City", "Nandurbar");
INSERT INTO Account (acc_no, branch_name, balance) VALUES
(101, 'Akurdi', 15000),
(102, 'Nigdi', 8000),
(103, 'Kharadi', 20000),
(104, 'Hadapsar', 12000),
(105, 'Viman Nagar', 5000);
INSERT INTO Account (accountNum, branchName, balance) VALUES
(2501, "Dhole Patil", 5000),
(2511, "Nagarwala", 1500),
(2521, "Peachtree", 2000),
(2512, "Bishops", 5000),
(2531, "Amanora", 1000);
INSERT INTO Customer (cust_name, cust_street, cust_city) VALUES
('Kalas', 'Street 1', 'Pune'),
('Mehul', 'Street 2', 'Pune'),
('Tanmay', 'Street 3', 'Pune'),
('Kshitij', 'Street 4', 'Pune'),
('Aditya', 'Street 5', 'Pune');
INSERT INTO Loan (loanNum, branchName, amount) VALUES
(155, "Dhole Patil", 500),
(156, "Nagarwala", 250),
(157, "Peachtree", 600),
(158, "Bishops", 1400),
(159, "Amanora", 25000);
INSERT INTO Depositor (cust_name, acc_no) VALUES
('Kalas', 101),
('Mehul', 102),
('Tanmay', 103),
('Kshitij', 101),
('Aditya', 104);
INSERT INTO Borrower VALUES
("Kalas", 156),
("Mehul", 158),
("Tanmay", 155),
("Kshitij", 157),
("Aditya", 159),
("Himanshu", 158);
INSERT INTO Loan (loan_no, branch_name, amount) VALUES
(201, 'Akurdi', 15000),
(202, 'Nigdi', 13000),
(203, 'Kharadi', 25000),
(204, 'Hadapsar', 18000),
(205, 'Akurdi', 1400);
INSERT INTO Depositor VALUES
("Kalas", 2511),
("Mehul", 2512),
("Tanmay", 2501),
("Kshitij", 2521),
("Aditya", 2531),
("Himanshu", 2512);
INSERT INTO Borrower (cust_name, loan_no) VALUES
('Mehul', 202),
('Tanmay', 203),
('Aditya', 205);
```
@ -91,14 +117,14 @@ INSERT INTO Depositor VALUES
1. Find the names of all branches in loan relation.
```sql
SELECT branchName FROM Loan;
SELECT DISTINCT branch_name FROM Loan;
```
2. Find all loan numbers for loans made at Akurdi Branch with loan amount > 12000.
```sql
SELECT loanNum FROM Loan WHERE amount>12000;
SELECT loan_no FROM Loan WHERE branch_name = 'Akurdi' AND amount > 12000;
```
@ -106,70 +132,73 @@ SELECT loanNum FROM Loan WHERE amount>12000;
```sql
SELECT * from Borrower;
SELECT Borrower.loanNum FROM Borrower INNER JOIN Loan ON Borrower.loanNum = Loan.loanNum;
SELECT Borrower.cust_name, Borrower.loan_no, Loan.amount FROM Borrower INNER JOIN Loan ON Borrower.loan_no = Loan.loan_no;
```
4. List all customers in alphabetical order who have loan from Akurdi branch.
```sql
SELECT customerName FROM Borrower INNER JOIN Loan ON Borrower.loanNum = Loan.loanNum WHERE branchName = "Akurdi" ORDER BY customerName;
SELECT cust_name FROM Borrower INNER JOIN Loan ON Borrower.loan_no = Loan.loan_no WHERE branch_name = 'Akurdi' ORDER BY cust_name;
```
5. Find all customers who have an account or loan or both at bank.
```sql
SELECT customerName FROM Depositor UNION SELECT customerName FROM Borrower;
SELECT cust_name FROM Depositor UNION SELECT cust_name FROM Borrower;
```
6. Find all customers who have both account and loan at bank.
```sql
SELECT customerName FROM Depositor INTERSECT SELECT customerName FROM Borrower;
SELECT cust_name FROM Depositor INTERSECT SELECT cust_name FROM Borrower;
```
7. Find all customers who have account but no loan at the bank.
```sql
SELECT customerName FROM Depositor WHERE customerName NOT IN (SELECT customerName FROM Borrower);
SELECT cust_name FROM Depositor WHERE cust_name NOT IN (SELECT cust_name FROM Borrower);
-- OR YOU CAN RUN THE BELOW COMMAND IF YOU ARE FEELING FANCY
-- SELECT Customer.cust_name FROM Customer INNER JOIN Depositor ON Customer.cust_name = Depositor.cust_name LEFT JOIN Borrower ON Customer.cust_name = Borrower.cust_name WHERE Borrower.cust_name IS NULL;
```
8. Find the average account balance at each branch
```sql
SELECT AVG(amount) FROM Loan;
SELECT branch_name, AVG(balance) FROM Account GROUP BY branch_name;
```
9. Find no. of depositors at each branch.
```sql
SELECT branchName, COUNT(*) AS noOfDepositors FROM Account JOIN Depositor ON Account.accountNum = Depositor.accountNum GROUP BY branchName;
SELECT branch_name, COUNT(*) FROM Account INNER JOIN Depositor ON Account.acc_no = Depositor.acc_no GROUP BY branch_name;
```
10. Find name of Customer and city where customer name starts with Letter K.
```sql
SELECT customerName, customerCity from Customer where customerName like "K%";
SELECT cust_name, cust_city FROM Customer WHERE cust_name LIKE 'K%';
```
11. Display distinct cities of branch.
```sql
SELECT DISTINCT branchName, branchCity FROM Branch;
SELECT DISTINCT branch_city FROM Branch;
```
12. Find the branches where average account balance > 1200
```sql
SELECT branchName FROM Account GROUP BY branchName HAVING AVG(balance) > 1200;
SELECT branch_name FROM Account GROUP BY branch_name HAVING AVG(balance) > 12000;
```
@ -183,26 +212,26 @@ SELECT COUNT(*) FROM Customer;
14. Calculate total loan amount given by bank.
```sql
SELECT SUM(amount) AS amount FROM Loan;
SELECT SUM(amount) FROM Loan;
```
15. Delete all loans with loan amount between 1300 and 1500.
```sql
DELETE FROM Borrower WHERE loanNum IN (SELECT loanNum FROM Loan WHERE amount > 1300 AND amount < 1500);
DELETE FROM Loan WHERE amount > 1300 AND amount < 1500;
DELETE FROM Borrower WHERE loan_no IN (SELECT loan_no FROM Loan WHERE amount BETWEEN 1300 AND 1500);
DELETE FROM Loan WHERE amount BETWEEN 1300 AND 1500;
```
16. Delete all tuples at every branch located in Nigdi.
```sql
DELETE FROM Borrower WHERE loanNum IN (SELECT loanNum FROM Loan WHERE branchName IN (SELECT branchName FROM Branch WHERE branchCity = "Nigdi"));
DELETE FROM Loan WHERE branchName = (SELECT branchName FROM Branch WHERE branchCity = "Nigdi");
DELETE FROM Depositor WHERE accountNum IN (SELECT accountNum FROM Account WHERE branchName IN (SELECT branchName FROM Branch WHERE branchCity = "Nigdi"));
DELETE FROM Account WHERE branchName = (SELECT branchName FROM Branch WHERE branchCity = "Nigdi");
DELETE FROM Branch WHERE branchName = “Nigdi”;
DELETE FROM Borrower WHERE loan_no IN (SELECT loan_no FROM Loan WHERE branch_name = 'Nigdi');
DELETE FROM Loan WHERE branch_name = 'Nigdi';
DELETE FROM Depositor WHERE acc_no IN (SELECT acc_no FROM Account WHERE branch_name = 'Nigdi');
DELETE FROM Account WHERE branch_name = 'Nigdi';
DELETE FROM Branch WHERE branch_name = 'Nigdi';
```