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Adding solutions to challenge 1 #57

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Adding solutions to challenge 1 #57

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42 changes: 34 additions & 8 deletions README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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">
Expand All @@ -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">
Expand All @@ -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">
Expand All @@ -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">
Expand All @@ -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');
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I think the only case when an ILIKE command should be used is when there's a pattern matching task to be done. Here I don't understand why a simple name = 'Colorado' wouldn't be enough.

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This is actually a mistake from my side, I was just trying things and I have never had uses ILIKE before, I was testing how it behaved and forgot to change it to the original name = 'Colorado', my apologies

```

<p align="center">
Expand All @@ -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">
Expand All @@ -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">
Expand All @@ -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
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nit: you could show this in a cleaner and more readable way

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