172 lines
3.6 KiB
Markdown
172 lines
3.6 KiB
Markdown
# S3 - SQL Queries (in MySQL)
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**Problem Statement:**
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Consider following Relation
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Account (Acc_no, branch_name,balance)
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Branch(branch_name,branch_city,assets)
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Customer(cust_name,cust_street,cust_city)
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Depositor(cust_name,acc_no)
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Loan(loan_no,branch_name,amount)
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Borrower(cust_name,loan_no)
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Create above tables with appropriate constraints like primary key,
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foreign key, not null etc.
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1. Find the branches where average account balance > 15000.
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2. Find number of tuples in customer relation.
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3. Calculate total loan amount given by bank.
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4. Delete all loans with loan amount between 1300 and 1500.
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5. Find the average account balance at each branch
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6. Find name of Customer and city where customer name starts with
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Letter P.
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---
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## Creating the database
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```sql
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CREATE DATABASE Bank3;
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USE Bank3;
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```
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## Creating tables:
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```sql
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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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## Declaring foreign keys
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```sql
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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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## Inserting data
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```sql
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INSERT INTO Branch VALUES
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('Wadia College', 'Pune', 50000),
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('PES', 'Pune', 65000),
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('Lohegaon', 'Pune', 350000),
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('Viman Nagar', 'Pune', 850000);
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INSERT INTO Customer VALUES
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('Kalas', 'Street 12', 'Pune'),
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('Himanshu', 'Street 15', 'Pune'),
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('Mehul', 'Street 29', 'Pune'),
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('Macho', 'Street 59', 'Mumbai'),
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('Gundeti', 'Street 40', 'Mumbai'),
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('Salvi', 'Street 8', 'Pune'),
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('Pintu', 'Street 55', 'Ahemadnagar'),
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('Piyush', 'Street 21', 'Assam');
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INSERT INTO Account VALUES
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(101, 'Lohegaon', 67000),
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(102, 'PES', 4324),
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(103, 'PES', 54670),
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(104, 'Viman Nagar', 5433),
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(105, 'Wadia College', 6462);
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INSERT INTO Depositor VALUES
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('Kalas', 101),
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('Macho', 104),
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('Gundeti', 105),
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('Salvi', 105);
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INSERT INTO Loan VALUES
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(201, 'Wadia College', 1800),
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(202, 'PES', 8500),
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(203, 'PES', 15000),
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(204, 'Wadia College', 5322),
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(205, 'Viman Nagar', 1300),
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(206, 'Lohegaon', 1450);
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INSERT INTO Borrower VALUES
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('Macho', 201),
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('Mehul', 202),
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('Himanshu', 203),
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('Salvi', 204);
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```
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## Queries
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1. Find the branches where average account balance > 15000.
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```sql
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SELECT branch_name FROM Account GROUP BY branch_name HAVING AVG(balance) > 15000 ;
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```
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2. Find number of tuples in customer relation.
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```sql
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SELECT COUNT(*) FROM Customer;
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```
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3. Calculate total loan amount given by bank.
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```sql
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SELECT SUM(amount) FROM Loan;
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```
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4. Delete all loans with loan amount between 1300 and 1500.
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```sql
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DELETE FROM Loan WHERE amount BETWEEN 1300 AND 1500;
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```
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5. Find the average account balance at each branch
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```sql
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SELECT branch_name, AVG(balance) FROM Account GROUP BY branch_name;
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```
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6. Find name of Customer and city where customer name starts with letter P.
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```sql
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SELECT cust_name, cust_city FROM Customer WHERE cust_name LIKE "P%";
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```
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---
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