https://github.com/shrunga92/restaurant_order_analysis_sql
This project is a structured SQL-based analysis of restaurant orders, aimed at deriving key insights from transactional data.
https://github.com/shrunga92/restaurant_order_analysis_sql
data-analysis sql
Last synced: 12 months ago
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This project is a structured SQL-based analysis of restaurant orders, aimed at deriving key insights from transactional data.
- Host: GitHub
- URL: https://github.com/shrunga92/restaurant_order_analysis_sql
- Owner: shrunga92
- Created: 2025-03-30T06:43:54.000Z (about 1 year ago)
- Default Branch: main
- Last Pushed: 2025-03-30T08:29:30.000Z (about 1 year ago)
- Last Synced: 2025-06-08T22:33:27.985Z (about 1 year ago)
- Topics: data-analysis, sql
- Homepage:
- Size: 120 KB
- Stars: 0
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
Awesome Lists containing this project
README
# Restaurant Order Analysis using SQL
This project is a structured SQL-based analysis of restaurant orders, aimed at deriving key insights from transactional data.
# Project Brief
## Database Schema
The project includes two tables:
# SQL Solutions
The SQL scripts in this repository cover:
Data Exploration: Understanding the structure and key statistics.
Revenue Analysis: Identifying top-selling and most profitable items.
Order Trends: Analyzing peak ordering times and customer preferences.
Performance Insights: Evaluating order frequency and restaurant efficiency.
# Key Findings:
The menu has 32 items across four categories: American, Asian, Mexican, and Italian.
Shrimp Scampi is the most expensive item, while Edamame is the least expensive.
The menu includes 9 Italian dishes, with Fettuccine Alfredo being the least expensive.
A total of 5,370 orders were placed, with 12,234 items ordered within the dataset's date range.
The highest number of items in a single order was Order #4305.
Hamburger was the most ordered item, while Chicken Tacos was the least ordered.
Order #440 had the highest spending, mostly on Italian dishes, suggesting demand for more Italian options.
The top 5 highest-spending orders were from Order IDs: 440, 2075, 1957, 330, and 2675.
These insights help optimize menu offerings, identify customer preferences, and improve restaurant sales strategies.
# Problems solved
### -> View the menu_items table and write a query to find the number of items on the menu
```sql
select * from menu_items;
select count(*) from menu_items;
```
### -> What are the price of least and most expensive items on the menu?
```sql
select min(price) from menu_items;
select max(price) from menu_items;
```
### -> How many Italian dishes are on the menu?
```sql
select count(*) from menu_items
where category = 'Italian';
```
### -> What are the least and most expensive Italian dishes on the menu? Sort the table.
```sql
select * from menu_items
where category = 'Italian'
order by price;
select * from menu_items
where category = 'Italian'
order by price desc;
```
### -> What are the items on the menu that have chicken in them.
```sql
select * from menu_items
where item_name like '%Chicken%';
```
### -> How many dishes are in each category?
```sql
select category , count(*)
from menu_items
group by 1;
```
### -> What is the average dish price within each category?
```sql
select category , avg(price) as avg_price
from menu_items
group by category;
```
### -> View the order_details table.
```sql
select * from order_details;
```
### -> What is the date range of the table?
```sql
SELECT min(order_date) , max(order_date)
from order_details;
```
### -> How many orders were made within this date range?
```sql
SELECT count(distinct order_id)
from order_details;
```
### -> How many items were ordered within this date range?
```sql
SELECT count(item_id)
from order_details;
```
### -> Which orders had the most number of items?
```sql
select order_id , count(item_id) as item_count
from order_details
GROUP by order_id
order by item_count desc;
```
### -> How many orders had more than 12 items?
```sql
select count(*) from
(select order_id , count(item_id) as item_count
from order_details
GROUP by order_id
HAVING count(item_id) > 12) as num_orders;
```
### -> Combine the menu_items and order_details tables into a single table
```sql
select * from order_details
left join menu_items on order_details.item_id = menu_items.menu_item_id;
```
### -> What were the least and most ordered items? What categories were they in?
```sql
select category , item_name , count(item_id) from order_details
left join menu_items on order_details.item_id = menu_items.menu_item_id
GROUP by 1,2
order by 3 desc ;
```
### -> What were the top 5 orders that spent the most money?
```sql
select order_id , sum(price)
from order_details
left join menu_items on order_details.item_id = menu_items.menu_item_id
where price is not null
group by 1
order by 2 DESC
limit 5;
```
### -> View the details of the highest spend order. Which specific items were purchased?
```sql
select order_id , item_name, sum(price)
from order_details
left join menu_items on order_details.item_id = menu_items.menu_item_id
where price is not null
group by 1 ,2
order by 3 DESC
limit 1;
```
### -> how many categories have max price below 15$
### solved using SUBQUERY
```sql
select count(*) from
(
select category , max(price)
from menu_items
group by 1
having max(price) < 15) as categories_having_max_price;
```
### solved using CTE.
```sql
with categories_having_max_price as (select category , max(price)
from menu_items
group by 1
having max(price) < 15)
select count(*) from categories_having_max_price;
```