Агрегирование подмножеств
Средний уровень SQL
Jasmin Ludolf
Data Science Content Developer, DataCamp
WHERE с агрегатными функциями
SELECT AVG(budget) AS avg_budget
FROM films
WHERE release_year >= 2010;
|avg_budget |
|--------------------|
|41072235.18324607...|
WHERE с агрегатными функциями
SELECT SUM(budget) AS sum_budget
FROM films
WHERE release_year = 2010;
|sum_budget|
|----------|
|8942365000|
SELECT MIN(budget) AS min_budget
FROM films
WHERE release_year = 2010;
|min_budget|
|----------|
|65000 |
WHERE с агрегатными функциями
SELECT MAX(budget) AS max_budget
FROM films
WHERE release_year = 2010;
|max_budget|
|----------|
|600000000 |
SELECT COUNT(budget) AS count_budget
FROM films
WHERE release_year = 2010;
|count_budget|
|------------|
|194 |
ROUND()
- Округление числа до заданного знака
SELECT AVG(budget) AS avg_budget
FROM films
WHERE release_year >= 2010;
|avg_budget |
|--------------------|
|41072235.18324607...|
ROUND(number_to_round, decimal_places)
SELECT ROUND(AVG(budget), 2) AS avg_budget
FROM films
WHERE release_year >= 2010;
|avg_budget |
|-----------|
|41072235.18|
ROUND() до целого числа
SELECT ROUND(AVG(budget)) AS avg_budget
FROM films
WHERE release_year >= 2010;
|avg_budget|
|----------|
|41072235 |
SELECT ROUND(AVG(budget), 0) AS avg_budget
FROM films
WHERE release_year >= 2010;
|avg_budget|
|----------|
|41072235 |
ROUND() с отрицательным параметром
SELECT ROUND(AVG(budget), -5) AS avg_budget
FROM films
WHERE release_year >= 2010;
|avg_budget|
|----------|
|41100000 |
- Только числовые поля
Давайте потренируемся!
Средний уровень SQL
