From bc0de03b81f0bf3f8d1b037b8bc06e8100cfcbae Mon Sep 17 00:00:00 2001
From: DavidH29 <54407010+DavidH29@users.noreply.github.com>
Date: Fri, 29 Nov 2024 15:06:07 -0600
Subject: [PATCH] Adding solutions to challenge 1

---
 README.md | 42 ++++++++++++++++++++++++++++++++++--------
 1 file changed, 34 insertions(+), 8 deletions(-)

diff --git a/README.md b/README.md
index d6594fa..0338da4 100644
--- a/README.md
+++ b/README.md
@@ -74,7 +74,9 @@ Now it's your turn to write SQL querys to achieve the following results:
 1. Count the total number of states in each country.
 
 ```
-Your query here
+select c.name, count(c.id) from countries c
+inner join states s on c.id = s.country_id 
+group by c.id;
 ```
 
 <p align="center">
@@ -84,7 +86,7 @@ Your query here
 2. How many employees do not have supervisores.
 
 ```
-Your query here
+select count(*) employees_without_bosses from employees e where supervisor_id is null;
 ```
 
 <p align="center">
@@ -94,7 +96,11 @@ Your query here
 3. List the top five offices address with the most amount of employees, order the result by country and display a column with a counter.
 
 ```
-Your query here
+select c.name, o.address, count(e.office_id) from offices o 
+inner join countries c on o.country_id = c.id 
+inner join employees e on o.id = e.office_id 
+group by o.id, c.name
+order by count(e.office_id) desc limit 5;
 ```
 
 <p align="center">
@@ -104,7 +110,9 @@ Your query here
 4. Three supervisors with the most amount of employees they are in charge.
 
 ```
-Your query here
+select supervisor_id, count(*) from employees e
+where supervisor_id is not null
+group by supervisor_id order by count(*) desc limit 3;
 ```
 
 <p align="center">
@@ -114,7 +122,8 @@ Your query here
 5. How many offices are in the state of Colorado (United States).
 
 ```
-Your query here
+select count(*) list_of_office from offices o 
+where state_id = (select id from states s where name ilike 'colorado');
 ```
 
 <p align="center">
@@ -124,7 +133,10 @@ Your query here
 6. The name of the office with its number of employees ordered in a desc.
 
 ```
-Your query here
+select o.name, count(e.office_id) from offices o  
+inner join employees e on o.id = e.office_id 
+group by o.id
+order by count(e.office_id) desc;
 ```
 
 <p align="center">
@@ -134,7 +146,14 @@ Your query here
 7. The office with more and less employees.
 
 ```
-Your query here
+(select o.address , count(e.office_id) from offices o  
+inner join employees e on o.id = e.office_id 
+group by o.id
+order by count(e.office_id) desc limit 1)
+union all (select o.address, count(e.office_id) from offices o  
+inner join employees e on o.id = e.office_id 
+group by o.id
+order by count(e.office_id) limit 1);
 ```
 
 <p align="center">
@@ -144,9 +163,16 @@ Your query here
 8. Show the uuid of the employee, first_name and lastname combined, email, job_title, the name of the office they belong to, the name of the country, the name of the state and the name of the boss (boss_name)
 
 ```
-Your query here
+select e.uuid, concat(e.first_name, ' ' ,e.last_name) full_name, e.email, e.job_title, o.name company, c.name country, s.name state, m.first_name boss_name
+from employees e
+inner join offices o on e.office_id = o.id
+inner join countries c on o.country_id = c.id
+inner join states s on o.state_id = s.id
+inner join employees m on e.supervisor_id = m.id;
 ```
 
 <p align="center">
  <img src="src/results/result8.png" alt="result_8"/>
 </p>
+
+Challenge 3 link: https://dbdiagram.io/d/EDR-Challenge-3-674a25dbe9daa85aca2d1246