https://github.com/nomercy10/mysql-zomato-design
A relational database design for applications like Zomato/Swiggy which will cover all their features
https://github.com/nomercy10/mysql-zomato-design
Last synced: 7 months ago
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A relational database design for applications like Zomato/Swiggy which will cover all their features
- Host: GitHub
- URL: https://github.com/nomercy10/mysql-zomato-design
- Owner: Nomercy10
- Created: 2021-03-29T09:12:13.000Z (over 4 years ago)
- Default Branch: master
- Last Pushed: 2021-03-29T12:23:03.000Z (over 4 years ago)
- Last Synced: 2025-02-04T13:28:54.653Z (8 months ago)
- Size: 6.84 KB
- Stars: 1
- Watchers: 2
- Forks: 1
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# Relational database design for applications like Zomato/Swiggy
> This is my try to design a relational database design for Zomato. I have used MySql for this as I am more comfortable with it
**Context**
- **FeaturesList.sql** : A sql file which will provide all the table schemas required for the applications like
- zomatomaindb
- customers
- restuarants
- zo_employee
- foods
- orders_details
- payments
- orders_food
- wallet
- **SQL queries**
### Data analysis using sql queries
`Count total number of users on Zomato:`
```sql
select count(*) as total_users from customers;
```
`Count total number of restuarants on Zomato:`
```sql
select count(*) as total_restaurants from restuarants;
```
`Count total number of employees of Zomato:`
```sql
select count(*) as total_employees from zo_employee;
```
`Find list of users who ordered food more than 2 times in a particular month:`
```sql
select cust_id, count(order_id) from orders_details
where cust_id in( select cust_id from orders_details where month(order_time) == 4)
group by cust_id
having count(order_id) > 2;
or
select c.cust_id, c.cust_name, count(o.order_id) as no_of_orders, month(o.order_time)
from customers c inner join orders_detail o on c.cust_id = o.cust_id
group by cust_id
having no_of_orders > 2
order by no_of_orders desc;
```
`Find top 3 delivery boy on the basis of customer rating and time to deliver the food item:`
*for eg on customer rating*
```sql
select o.emp_id, z.emp_name, z.emp_rating, timestampdiff(minute, o.order_time,o.delivered_time) as delivered_time_minute
from orders_details o inner join zo_employee on o.emp_id = z.emp_id
group by emp_id
order by z.emp_rating desc
limit 3;
```
*for eg on delivery time*
```sql
select o.emp_id, z.emp_name, z.emp_rating, timestampdiff(minute, o.order_time,o.delivered_time) as delivered_time_minute
from orders_details o inner join zo_employee on o.emp_id = z.emp_id
group by emp_id
order by delivered_time_minute asc
limit 3;
```
*for both the cases*
```sql
select o.emp_id, z.emp_name, z.emp_rating, timestampdiff(minute, o.order_time,o.delivered_time) as delivered_time_minute
from orders_details o inner join zo_employee on o.emp_id = z.emp_id
group by emp_id
order by delivered_time_minute, z.emp_rating asc
limit 3;
```
`Find users who order food from same restuarants more than 2 times in a week:`
```sql
select o.cust_id, c.cust_name, rest_id, count(*) as orders, week(order_time)
from orders_details inner join customer c on o.cust_id == c.cust_id
group by rest_id
having orders > 2;
```
### ER diagram
