SQL 速查表(MySQL, PostgreSQL)

    mysql -h localhost -u root -p
// /usr/local/Cellar/mysql@xx/x.xx.xx/bin/mysql -h localhost -u root
  

    mysql -u username -p database_name < file.sql
  

    mysqldump --user root --password  --all-databases > all-databases.sql
mysql --user root --password mysql < all-databases.sql
  

    show databases;
  

    use [db name];
show tables;
  

    SELECT User, Host, authentication_string FROM mysql.user;
  

    # Create a user
CREATE USER 'db_user'@'localhost' IDENTIFIED BY 'password';

# Create a database
CREATE DATABASE example_db;

# Grant privileges
GRANT ALL ON example_db.* TO 'db_user'@'localhost';
  

    # Delete database
DROP DATABASE example_db;

# Delete user
DROP USER 'db_user'@'localhost';
  

    quit
  

    SELECT @@GLOBAL.sql_mode;
SELECT @@SESSION.sql_mode;

# result: 'ONLY_FULL_GROUP_BY,STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION'

SET GLOBAL sql_mode = '...';
SET SESSION sql_mode = '...';

# Remove ONLY_FULL_GROUP_BY
SET GLOBAL sql_mode=(SELECT REPLACE(@@sql_mode,'ONLY_FULL_GROUP_BY',''));
  

    INNER JOIN：如果表中有至少一个匹配，则返回行
LEFT JOIN：即使右表中没有匹配，也从左表返回所有的行
RIGHT JOIN：即使左表中没有匹配，也从右表返回所有的行
FULL JOIN：只要其中一个表中存在匹配，则返回行
# LEFT JOIN and LEFT OUTER JOIN are equivalent.
  

    select uuid_short() as id;
  

    USE database_name;
SELECT COUNT(*) FROM table_name WHERE table_name.a_id NOT IN (1,2,3,4) AND table_name.b_id = 12 AND table_name.c_id IN (SELECT some_id FROM another_table WHERE another_table.d_id = 34);