ExposedKotlin 的轻量级 ORM
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。
目前支持的数据库方言：
PostgreSQL
MySQL
SQLite
H2
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评论