feat: adding solutions of the exercises - challenge 1

README.md

@@ -74,7 +74,11 @@ 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(s.name) as "count"
+FROM states s
+INNER JOIN countries c ON s.country_id = c.id
+GROUP BY c.name;
 ```
 
 <p align="center">
@@ -84,7 +88,10 @@ Your query here
 2. How many employees do not have supervisores.
 
 ```
-Your query here
+SELECT COUNT(*)
+	employees_without_bosses
+FROM employees
+WHERE supervisor_id IS NULL;
 ```
 
 <p align="center">
@@ -94,7 +101,16 @@ 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.uuid) AS "count"
+FROM employees e
+INNER JOIN offices o ON e.office_id = o.id
+INNER JOIN countries c ON o.country_id = c.id
+GROUP BY c.name, o.address
+ORDER BY "count" DESC
+LIMIT 5;
 ```
 
 <p align="center">
@@ -104,7 +120,12 @@ Your query here
 4. Three supervisors with the most amount of employees they are in charge.
 
 ```
-Your query here
+SELECT e.supervisor_id, COUNT(*) AS total_employees
+FROM employees e
+WHERE e.supervisor_id IS NOT NULL
+GROUP BY e.supervisor_id
+ORDER BY total_employees DESC
+LIMIT 3;
 ```
 
 <p align="center">
@@ -114,7 +135,10 @@ Your query here
 5. How many offices are in the state of Colorado (United States).
 
 ```
-Your query here
+SELECT
+	COUNT(*) AS list_of_office
+FROM offices o
+WHERE o.country_id = 1 AND o.state_id = 8;
 ```
 
 <p align="center">
@@ -124,7 +148,13 @@ Your query here
 6. The name of the office with its number of employees ordered in a desc.
 
 ```
-Your query here
+SELECT
+	o.name AS "name",
+	COUNT(e.office_id) AS "count"
+FROM employees e
+INNER JOIN offices o ON e.office_id = o.id
+GROUP BY o.name
+ORDER BY "count" DESC;
 ```
 
 <p align="center">
@@ -134,7 +164,19 @@ Your query here
 7. The office with more and less employees.
 
 ```
-Your query here
+WITH cte AS (
+  SELECT
+    o.name AS "name",
+    COUNT(e.office_id) AS "count",
+    ROW_NUMBER() OVER (ORDER BY COUNT(e.office_id) DESC) AS rn_desc,
+    ROW_NUMBER() OVER (ORDER BY COUNT(e.office_id) ASC) AS rn_asc
+  FROM employees e
+  INNER JOIN offices o ON e.office_id = o.id
+  GROUP BY o.name
+);
+SELECT "name", "count"
+FROM cte
+WHERE rn_asc = 1 OR rn_desc = 1;
 ```
 
 <p align="center">
@@ -144,7 +186,18 @@ 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) AS full_name,
+    e.email,
+    e.job_title,
+    o.name AS office_name,
+    co.name AS country_name,
+    s.name AS state_name
+FROM employees e
+INNER JOIN offices o ON e.office_id = o.id
+INNER JOIN states s ON o.state_id = s.id
+INNER JOIN countries co ON s.country_id = co.id;
 ```
 
 <p align="center">

docker-compose.yml

@@ -0,0 +1,16 @@
+version: "3.9"
+services:
+  postgres:
+    image: postgres:latest
+    container_name: nerdery-container
+    environment:
+      POSTGRES_USER: postgres
+      POSTGRES_PASSWORD: password123
+    ports:
+      - "5000:5432"
+    volumes:
+      - postgres_data:/var/lib/postgresql/data
+      - ./src/dump.sql:/docker-entrypoint-initdb.d/dump.sql
+
+volumes:
+  postgres_data:

src/answers.sql

@@ -1,15 +1,73 @@
 -- Your answers here:
 -- 1
-
+SELECT 
+    c.name, COUNT(s.name) as "count" 
+FROM states s
+INNER JOIN countries c ON s.country_id = c.id
+GROUP BY c.name;
 -- 2
-
+SELECT COUNT(*) 
+	employees_without_bosses 
+FROM employees 
+WHERE supervisor_id IS NULL;
 -- 3
-
+SELECT 
+    c.name, 
+    o.address,
+	COUNT(e.uuid) AS "count"
+FROM employees e
+INNER JOIN offices o ON e.office_id = o.id
+INNER JOIN countries c ON o.country_id = c.id
+GROUP BY c.name, o.address
+ORDER BY "count" DESC
+LIMIT 5;
 -- 4
-
+SELECT e.supervisor_id, COUNT(*) AS total_employees
+FROM employees e
+WHERE e.supervisor_id IS NOT NULL
+GROUP BY e.supervisor_id
+ORDER BY total_employees DESC
+LIMIT 3;
 -- 5
-
+SELECT 
+	COUNT(*) AS list_of_office
+FROM offices o
+WHERE o.country_id = 1 AND o.state_id = 8;
 -- 6
-
+SELECT 
+	o.name AS "name",
+	COUNT(e.office_id) AS "count"
+FROM employees e
+INNER JOIN offices o ON e.office_id = o.id
+GROUP BY o.name
+ORDER BY "count" DESC;
 -- 7
-
+WITH cte AS (
+  SELECT 
+    o.name AS "name",
+    COUNT(e.office_id) AS "count",
+    ROW_NUMBER() OVER (ORDER BY COUNT(e.office_id) DESC) AS rn_desc,
+    ROW_NUMBER() OVER (ORDER BY COUNT(e.office_id) ASC) AS rn_asc
+  FROM employees e
+  INNER JOIN offices o ON e.office_id = o.id
+  GROUP BY o.name
+);
+SELECT "name", "count"
+FROM cte
+WHERE rn_asc = 1 OR rn_desc = 1;
+-- 8
+SELECT 
+    e.uuid,
+    CONCAT(e.first_name, ' ', e.last_name) AS full_name,
+    e.email,
+    e.job_title,
+    o.name AS office_name,
+    co.name AS country_name,
+    s.name AS state_name,
+    supervisors.first_name boss_name 
+FROM employees e
+INNER JOIN offices o ON e.office_id = o.id
+INNER JOIN states s ON o.state_id = s.id
+INNER JOIN countries co ON s.country_id = co.id;
+INNER JOIN employees supervisors ON e.supervisor_id = supervisors.id
+WHERE supervisors.first_name IS NOT NULL;