SQL Intermediário com IA
Jasmin Ludolf
Senior Data Science Content Developer
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SELECT DISTINCT language
FROM films;
|language|
|--------|
|Danish |
|Greek |
|Dzongkha|
|... |
SELECT language
FROM films
GROUP BY language;
|language|
|--------|
|Danish |
|Greek |
|Dzongkha|
|... |
Prompt: Agrupar filmes por idioma
SELECT language, COUNT(*) AS lang_count
FROM films
GROUP BY language;
|language|lang_count|
|--------|----------|
| |11 |
|Arabic |5 |
|Kannada |1 |
|Czech |1 |
...
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COUNT()Prompt: Mostrar todos os títulos de filmes agrupados por idioma
SELECT title
FROM films
GROUP BY language;
column "films.title" must appear in the
GROUP BY clause or be used in an
aggregate function LINE 1...
Correções de IA:
COUNT()SELECT language, COUNT(title) AS count
FROM films
GROUP BY language;
SELECT title, language
FROM films
GROUP BY title, language;
Prompt: Contar filmes por país e idioma
SELECT country, language, COUNT(*) AS film_count
FROM films
GROUP BY country, language;
|country |language|film_count|
|------------------|--------|----------|
| | |1 |
|Pakistan |Urdu |1 |
|Mexico |English |3 |
|Dominican Republic|Spanish |1 |
...
Prompt: Classificar idiomas por número de filmes, mostrando os 3 primeiros
SELECT language, COUNT(*) AS film_count
FROM films
GROUP BY language
ORDER BY film_count DESC
LIMIT 3;
|language|film_count|
|--------|----------|
|English |4635 |
|French |72 |
|Spanish |40 |
SQL Intermediário com IA