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81 lines
4.6 KiB
81 lines
4.6 KiB
# Intermediate SQL
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## Lesson Objectives tools1. Alter a table
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1. Alter a table
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1. Limit
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1. Sorting
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1. Aggregation
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1. Joins
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## Alter a table
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```sql
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ALTER TABLE users ADD COLUMN test VARCHAR(20); -- add an test string column
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ALTER TABLE users DROP COLUMN test; -- drop the test column
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ALTER TABLE users RENAME name TO first_name -- rename a column
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ALTER TABLE users ADD COLUMN id serial PRIMARY KEY; -- add an id column that increments with each new row
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ALTER TABLE users RENAME TO people; -- rename a table
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ALTER TABLE people ALTER COLUMN first_name TYPE text; -- chang the data type of a column
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```
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## Limit
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```sql
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SELECT * FROM people LIMIT 1; -- select all rows from people table, but show only the first column
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SELECT * FROM people LIMIT 1 OFFSET 1; -- select all rows from people table, but show only one column. Skip the first row
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```
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## Sorting
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```sql
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SELECT * FROM people ORDER BY first_name ASC; -- select all rows from people table, order by name alphabetically
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SELECT * FROM people ORDER BY first_name DESC; -- select all rows from people table, order by name reverse alphabetically
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SELECT * FROM people ORDER BY age ASC; -- select all rows from people table, order by age ascending
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SELECT * FROM people ORDER BY age DESC; -- select all rows from people table, order by age descending
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```
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## Aggregation
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```sql
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SELECT SUM(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name. Show the SUM of the ages of each group. Also show what name each group has
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SELECT AVG(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name. Show the AVG of the ages of each group. Also show what name each group has
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SELECT MIN(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name. Show the MAX of the ages of each group. Also show what name each group has
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SELECT MAX(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name. Show the MIN of the ages of each group. Also show what name each group has
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SELECT COUNT(age), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name. Show how many rows have a value in the age column. Also show what name each group has
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SELECT COUNT(*), first_name FROM people GROUP BY first_name; -- divide all rows into groups by name. Show the number of rows in each group. Also show what name each group has
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SELECT array_agg(first_name), age FROM people GROUP BY age; -- divide all rows into groups by age. List the names in each group and show what age each group has
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```
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## JOINS
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```sql
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SELECT * FROM people CROSS JOIN companies;
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SELECT * FROM people JOIN companies ON people.employer_id = companies.id -- find all people who have an employer_id column set and show which company they work for
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SELECT * FROM people LEFT JOIN companies ON people.employer_id = companies.id -- find all people have an employer_id column set and show which company they work for. In addition to this set, add on all people who do not have an employer_id column set
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SELECT * FROM people RIGHT JOIN companies ON people.employer_id = companies.id -- find all people have an employer_id column set and show which company they work for. In addition to this set, add on all companies who do not have any people with employer_id columns set to the company's id column
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SELECT * FROM people FULL OUTER JOIN companies ON people.employer_id = companies.id; -- find all people have an employer_id column set and show which company they work for. In addition to this set, add on all companies who do not have any people with employer_id columns set to the company's id column and all people who do not have an employer_id column set
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```
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## Combining Statments
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You can combine `WHERE`, `LIMIT`, `ORDER BY`, `OFFSET` with your `JOIN` statements:
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```sql
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SELECT * FROM people LEFT JOIN companies ON people.employer_id = companies.id WHERE first_name LIKE 'M%' ORDER BY age ASC LIMIT 2 OFFSET 1;
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```
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You can even add `GROUP BY` and aggregation functions too:
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```sql
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SELECT AVG(age), first_name FROM people LEFT JOIN companies ON people.employer_id = companies.id WHERE first_name LIKE 'M%' GROUP BY first_name ORDER BY avg ASC LIMIT 2 OFFSET 1;
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```
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The order where you add these in the statement matters. The following breaks (`GROUP BY` must come before `ORDER BY`, `LIMIT`, `OFFSET`):
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```sql
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SELECT AVG(age), first_name FROM people LEFT JOIN companies ON people.employer_id = companies.id WHERE first_name LIKE 'M%' ORDER BY avg ASC LIMIT 2 OFFSET 1 GROUP BY first_name;
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```
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If it breaks, just try reordering until it works. You can also research the order in which they need to come in the statement
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