Simplified queries for A2:
- 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
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171
Practical/Assignment-A2/Queries-A2.md
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171
Practical/Assignment-A2/Queries-A2.md
Executable file → Normal file
@ -13,74 +13,100 @@ USE Database_A2;
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## Creating tables and specify primary keys:
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```sql
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CREATE TABLE Account(accountNum INT, branchName VARCHAR(50), balance INT, PRIMARY KEY (accountNum));
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CREATE TABLE Branch(branchName VARCHAR(50), branchCity VARCHAR(50), assets INT, PRIMARY KEY (branchName));
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CREATE TABLE Customer (customerName VARCHAR(50), customerStreet VARCHAR(50), customerCity VARCHAR(50), PRIMARY KEY (customerName));
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CREATE TABLE Depositor (customerName VARCHAR(50), accountNum INT);
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CREATE TABLE Loan (loanNum INT, branchName VARCHAR(50), amount INT, PRIMARY KEY (loanNum));
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CREATE TABLE Borrower (customerName VARCHAR(50), loanNum INT);
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CREATE TABLE Account (
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acc_no INT,
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branch_name VARCHAR(255),
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balance INT,
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PRIMARY KEY (acc_no)
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);
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CREATE TABLE Branch (
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branch_name VARCHAR(255),
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branch_city VARCHAR(255),
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assets INT,
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PRIMARY KEY (branch_name)
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);
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CREATE TABLE Customer (
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cust_name VARCHAR(255),
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cust_street VARCHAR(255),
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cust_city VARCHAR(255),
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PRIMARY KEY (cust_name)
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);
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CREATE TABLE Depositor (
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cust_name VARCHAR(255),
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acc_no INT
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);
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CREATE TABLE Loan (
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loan_no INT,
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branch_name VARCHAR(255),
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amount INT,
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PRIMARY KEY (loan_no)
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);
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CREATE TABLE Borrower (
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cust_name VARCHAR(255),
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loan_no INT
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);
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```
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## Altering tables to specify foreign keys
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```sql
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ALTER TABLE Account ADD FOREIGN KEY (branchName) REFERENCES Branch(branchName);
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ALTER TABLE Depositor ADD FOREIGN KEY (customerName) REFERENCES Customer (customerName);
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ALTER TABLE Depositor ADD FOREIGN KEY (accountNum) REFERENCES Account (accountNum);
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ALTER TABLE Loan ADD FOREIGN KEY (branchName) REFERENCES Branch (branchName);
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ALTER TABLE Borrower ADD FOREIGN KEY (customerName) REFERENCES Customer (customerName);
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ALTER TABLE Borrower ADD FOREIGN KEY (loanNum) REFERENCES Loan (loanNum);
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ALTER TABLE Account ADD FOREIGN KEY (branch_name) REFERENCES Branch (branch_name);
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ALTER TABLE Depositor ADD FOREIGN KEY (cust_name) REFERENCES Customer (cust_name);
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ALTER TABLE Depositor ADD FOREIGN KEY (acc_no) REFERENCES Account (acc_no);
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ALTER TABLE Loan ADD FOREIGN KEY (branch_name) REFERENCES Branch (branch_name);
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ALTER TABLE Borrower ADD FOREIGN KEY (cust_name) REFERENCES Customer (cust_name);
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ALTER TABLE Borrower ADD FOREIGN KEY (loan_no) REFERENCES Loan (loan_no);
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```
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## Adding data
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```sql
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INSERT INTO Branch (branchName, branchCity, assets) VALUES
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("Dhole Patil", "Kharadi", 50000),
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("Nagarwala", "Akurdi", 20000),
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("Peachtree", "Wakad", 35000),
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("Bishops", "Nigdi", 10000),
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("Amanora", "Hadapsar", 60000);
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INSERT INTO Branch (branch_name, branch_city, assets) VALUES
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('Akurdi', 'Pune', 5000000),
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('Nigdi', 'Pune', 3000000),
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('Kharadi', 'Pune', 4000000),
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('Hadapsar', 'Pune', 3500000),
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('Viman Nagar', 'Pune', 4500000);
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INSERT INTO Customer (customerName, customerStreet, customerCity) VALUES
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("Kalas", "Airport Road", "Pune"),
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("Mehul", "Shahdha", "Nandurbar"),
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("Tanmay", "Porwal Road", "Pune"),
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("Kshitij", "Hadapasar", "Pune"),
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("Aditya", "Mira RD", "Mumbai"),
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("Himanshu", "Smart City", "Nandurbar");
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INSERT INTO Account (acc_no, branch_name, balance) VALUES
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(101, 'Akurdi', 15000),
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(102, 'Nigdi', 8000),
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(103, 'Kharadi', 20000),
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(104, 'Hadapsar', 12000),
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(105, 'Viman Nagar', 5000);
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INSERT INTO Account (accountNum, branchName, balance) VALUES
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(2501, "Dhole Patil", 5000),
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(2511, "Nagarwala", 1500),
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(2521, "Peachtree", 2000),
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(2512, "Bishops", 5000),
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(2531, "Amanora", 1000);
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INSERT INTO Customer (cust_name, cust_street, cust_city) VALUES
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('Kalas', 'Street 1', 'Pune'),
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('Mehul', 'Street 2', 'Pune'),
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('Tanmay', 'Street 3', 'Pune'),
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('Kshitij', 'Street 4', 'Pune'),
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('Aditya', 'Street 5', 'Pune');
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INSERT INTO Loan (loanNum, branchName, amount) VALUES
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(155, "Dhole Patil", 500),
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(156, "Nagarwala", 250),
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(157, "Peachtree", 600),
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(158, "Bishops", 1400),
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(159, "Amanora", 25000);
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INSERT INTO Depositor (cust_name, acc_no) VALUES
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('Kalas', 101),
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('Mehul', 102),
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('Tanmay', 103),
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('Kshitij', 101),
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('Aditya', 104);
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INSERT INTO Borrower VALUES
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("Kalas", 156),
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("Mehul", 158),
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("Tanmay", 155),
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("Kshitij", 157),
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("Aditya", 159),
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("Himanshu", 158);
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INSERT INTO Loan (loan_no, branch_name, amount) VALUES
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(201, 'Akurdi', 15000),
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(202, 'Nigdi', 13000),
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(203, 'Kharadi', 25000),
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(204, 'Hadapsar', 18000),
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(205, 'Akurdi', 1400);
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INSERT INTO Depositor VALUES
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("Kalas", 2511),
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("Mehul", 2512),
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("Tanmay", 2501),
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("Kshitij", 2521),
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("Aditya", 2531),
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("Himanshu", 2512);
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INSERT INTO Borrower (cust_name, loan_no) VALUES
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('Mehul', 202),
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('Tanmay', 203),
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('Aditya', 205);
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```
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@ -91,14 +117,14 @@ INSERT INTO Depositor VALUES
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1. Find the names of all branches in loan relation.
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```sql
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SELECT branchName FROM Loan;
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SELECT DISTINCT branch_name FROM Loan;
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```
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2. Find all loan numbers for loans made at Akurdi Branch with loan amount > 12000.
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```sql
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SELECT loanNum FROM Loan WHERE amount>12000;
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SELECT loan_no FROM Loan WHERE branch_name = 'Akurdi' AND amount > 12000;
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```
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@ -106,70 +132,73 @@ SELECT loanNum FROM Loan WHERE amount>12000;
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```sql
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SELECT * from Borrower;
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SELECT Borrower.loanNum FROM Borrower INNER JOIN Loan ON Borrower.loanNum = Loan.loanNum;
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SELECT Borrower.cust_name, Borrower.loan_no, Loan.amount FROM Borrower INNER JOIN Loan ON Borrower.loan_no = Loan.loan_no;
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```
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4. List all customers in alphabetical order who have loan from Akurdi branch.
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```sql
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SELECT customerName FROM Borrower INNER JOIN Loan ON Borrower.loanNum = Loan.loanNum WHERE branchName = "Akurdi" ORDER BY customerName;
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SELECT cust_name FROM Borrower INNER JOIN Loan ON Borrower.loan_no = Loan.loan_no WHERE branch_name = 'Akurdi' ORDER BY cust_name;
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```
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5. Find all customers who have an account or loan or both at bank.
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```sql
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SELECT customerName FROM Depositor UNION SELECT customerName FROM Borrower;
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SELECT cust_name FROM Depositor UNION SELECT cust_name FROM Borrower;
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```
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6. Find all customers who have both account and loan at bank.
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```sql
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SELECT customerName FROM Depositor INTERSECT SELECT customerName FROM Borrower;
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SELECT cust_name FROM Depositor INTERSECT SELECT cust_name FROM Borrower;
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```
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7. Find all customers who have account but no loan at the bank.
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```sql
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SELECT customerName FROM Depositor WHERE customerName NOT IN (SELECT customerName FROM Borrower);
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SELECT cust_name FROM Depositor WHERE cust_name NOT IN (SELECT cust_name FROM Borrower);
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-- OR YOU CAN RUN THE BELOW COMMAND IF YOU ARE FEELING FANCY
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-- 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;
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```
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8. Find the average account balance at each branch
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```sql
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SELECT AVG(amount) FROM Loan;
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SELECT branch_name, AVG(balance) FROM Account GROUP BY branch_name;
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```
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9. Find no. of depositors at each branch.
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```sql
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SELECT branchName, COUNT(*) AS noOfDepositors FROM Account JOIN Depositor ON Account.accountNum = Depositor.accountNum GROUP BY branchName;
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SELECT branch_name, COUNT(*) FROM Account INNER JOIN Depositor ON Account.acc_no = Depositor.acc_no GROUP BY branch_name;
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```
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10. Find name of Customer and city where customer name starts with Letter K.
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```sql
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SELECT customerName, customerCity from Customer where customerName like "K%";
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SELECT cust_name, cust_city FROM Customer WHERE cust_name LIKE 'K%';
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```
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11. Display distinct cities of branch.
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```sql
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SELECT DISTINCT branchName, branchCity FROM Branch;
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SELECT DISTINCT branch_city FROM Branch;
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```
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12. Find the branches where average account balance > 1200
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```sql
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SELECT branchName FROM Account GROUP BY branchName HAVING AVG(balance) > 1200;
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SELECT branch_name FROM Account GROUP BY branch_name HAVING AVG(balance) > 12000;
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```
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@ -183,26 +212,26 @@ SELECT COUNT(*) FROM Customer;
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14. Calculate total loan amount given by bank.
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```sql
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SELECT SUM(amount) AS amount FROM Loan;
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SELECT SUM(amount) FROM Loan;
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```
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15. Delete all loans with loan amount between 1300 and 1500.
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```sql
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DELETE FROM Borrower WHERE loanNum IN (SELECT loanNum FROM Loan WHERE amount > 1300 AND amount < 1500);
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DELETE FROM Loan WHERE amount > 1300 AND amount < 1500;
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DELETE FROM Borrower WHERE loan_no IN (SELECT loan_no FROM Loan WHERE amount BETWEEN 1300 AND 1500);
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DELETE FROM Loan WHERE amount BETWEEN 1300 AND 1500;
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```
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16. Delete all tuples at every branch located in Nigdi.
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```sql
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DELETE FROM Borrower WHERE loanNum IN (SELECT loanNum FROM Loan WHERE branchName IN (SELECT branchName FROM Branch WHERE branchCity = "Nigdi"));
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DELETE FROM Loan WHERE branchName = (SELECT branchName FROM Branch WHERE branchCity = "Nigdi");
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DELETE FROM Depositor WHERE accountNum IN (SELECT accountNum FROM Account WHERE branchName IN (SELECT branchName FROM Branch WHERE branchCity = "Nigdi"));
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DELETE FROM Account WHERE branchName = (SELECT branchName FROM Branch WHERE branchCity = "Nigdi");
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DELETE FROM Branch WHERE branchName = “Nigdi”;
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DELETE FROM Borrower WHERE loan_no IN (SELECT loan_no FROM Loan WHERE branch_name = 'Nigdi');
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DELETE FROM Loan WHERE branch_name = 'Nigdi';
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DELETE FROM Depositor WHERE acc_no IN (SELECT acc_no FROM Account WHERE branch_name = 'Nigdi');
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DELETE FROM Account WHERE branch_name = 'Nigdi';
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DELETE FROM Branch WHERE branch_name = 'Nigdi';
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```
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