Exposed 介绍
Exposed 是 Kotlin 实现的轻量级 ORM，也是 JetBrains 和 Kotlin 官方倾力推荐的 ORM。
该 ORM 十分轻量级，只做了底层封装、Dao 基础和 Model 基础，中上层的封装完全没有，可嵌入 Spring。
推荐一个 starter 项目：Kotlin + Ktor +
Exposed
示例代码
sql DSL 示例：
import org.jetbrains.exposed.sql.* import org.jetbrains.exposed.sql.transactions.transaction object Users : Table() { val id = varchar("id", 10).primaryKey() // Column<String> val name = varchar("name", length = 50) // Column<String> val cityId = (integer("city_id") references Cities.id).nullable() // Column<Int?> } object Cities : Table() { val id = integer("id").autoIncrement().primaryKey() // Column<Int> val name = varchar("name", 50) // Column<String> } fun main(args: Array<String>) { Database.connect("jdbc:h2:mem:test", driver = "org.h2.Driver") transaction { SchemaUtils.create (Cities, Users) val saintPetersburgId = Cities.insert { it[name] = "St. Petersburg" } get Cities.id val munichId = Cities.insert { it[name] = "Munich" } get Cities.id Cities.insert { it[name] = "Prague" } Users.insert { it[id] = "andrey" it[name] = "Andrey" it[cityId] = saintPetersburgId } Users.insert { it[id] = "sergey" it[name] = "Sergey" it[cityId] = munichId } Users.insert { it[id] = "eugene" it[name] = "Eugene" it[cityId] = munichId } Users.insert { it[id] = "alex" it[name] = "Alex" it[cityId] = null } Users.insert { it[id] = "smth" it[name] = "Something" it[cityId] = null } Users.update({Users.id eq "alex"}) { it[name] = "Alexey" } Users.deleteWhere{Users.name like "%thing"} println("All cities:") for (city in Cities.selectAll()) { println("${city[Cities.id]}: ${city[Cities.name]}") } println("Manual join:") (Users innerJoin Cities).slice(Users.name, Cities.name). select {(Users.id.eq("andrey") or Users.name.eq("Sergey")) and Users.id.eq("sergey") and Users.cityId.eq(Cities.id)}.forEach { println("${it[Users.name]} lives in ${it[Cities.name]}") } println("Join with foreign key:") (Users innerJoin Cities).slice(Users.name, Users.cityId, Cities.name). select {Cities.name.eq("St. Petersburg") or Users.cityId.isNull()}.forEach { if (it[Users.cityId] != null) { println("${it[Users.name]} lives in ${it[Cities.name]}") } else { println("${it[Users.name]} lives Nowhere") } } println("Functions and group by:") ((Cities innerJoin Users).slice(Cities.name, Users.id.count()).selectAll().groupBy(Cities.name)).forEach { val cityName = it[Cities.name] val userCount = it[Users.id.count()] if (userCount > 0) { println("$userCount user(s) live(s) in $cityName") } else { println("Nobody lives in $cityName") } } SchemaUtils.drop (Users, Cities) } }
Outputs:
sql: CREATE TABLE IF NOT EXISTS Cities (id INT AUTO_INCREMENT NOT NULL, name VARCHAR(50) NOT NULL, CONSTRAINT pk_Cities PRIMARY KEY (id)) sql: CREATE TABLE IF NOT EXISTS Users (id VARCHAR(10) NOT NULL, name VARCHAR(50) NOT NULL, city_id INT NULL, CONSTRAINT pk_Users PRIMARY KEY (id)) sql: ALTER TABLE Users ADD FOREIGN KEY (city_id) REFERENCES Cities(id) sql: INSERT INTO Cities (name) VALUES ('St. Petersburg') sql: INSERT INTO Cities (name) VALUES ('Munich') sql: INSERT INTO Cities (name) VALUES ('Prague') sql: INSERT INTO Users (id, name, city_id) VALUES ('andrey', 'Andrey', 1) sql: INSERT INTO Users (id, name, city_id) VALUES ('sergey', 'Sergey', 2) sql: INSERT INTO Users (id, name, city_id) VALUES ('eugene', 'Eugene', 2) sql: INSERT INTO Users (id, name, city_id) VALUES ('alex', 'Alex', NULL) sql: INSERT INTO Users (id, name, city_id) VALUES ('smth', 'Something', NULL) sql: UPDATE Users SET name='Alexey' WHERE Users.id = 'alex' sql: DELETE FROM Users WHERE Users.name LIKE '%thing' All cities: sql: SELECT Cities.id, Cities.name FROM Cities 1: St. Petersburg 2: Munich 3: Prague Manual join: sql: SELECT Users.name, Cities.name FROM Users INNER JOIN Cities ON Cities.id = Users.city_id WHERE ((Users.id = 'andrey') or (Users.name = 'Sergey')) and Users.id = 'sergey' and Users.city_id = Cities.id Sergey lives in Munich Join with foreign key: sql: SELECT Users.name, Users.city_id, Cities.name FROM Users INNER JOIN Cities ON Cities.id = Users.city_id WHERE (Cities.name = 'St. Petersburg') or (Users.city_id IS NULL) Andrey lives in St. Petersburg Functions and group by: sql: SELECT Cities.name, COUNT(Users.id) FROM Cities INNER JOIN Users ON Cities.id = Users.city_id GROUP BY Cities.name 1 user(s) live(s) in St. Petersburg 2 user(s) live(s) in Munich sql: DROP TABLE Users sql: DROP TABLE Cities