Is there a db connection pool for swift in perfect?

I am looking for a database connection pool wrapper for perfect (swift 3 or above). I couldn't find any through search so I decided to ask if someone could provide a hint.

Thanks in advance

I would rather say high efficient re-use connection other than a pool. Take Perfect-MySQL example:

import MySQL
import PerfectThread
#if os(Linux)
import Glibc
#else
import Darwin
#endif

let mysql = MySQL()
let lock = Threading.Lock()
var jobs = 10

func now(_ id: Int) {
  print("Job Now #", id)
  lock.doWithLock {
    let x = mysql.query(statement: "SELECT now() as time")
    guard x, let y = mysql.storeResults(),
    let row = y.next() else {
      print(mysql.errorMessage())
      return
    }
    print(row[0] ?? "Now() FAILED")
    y.close()
    jobs -= 1
  }
}

func user(_ id: Int) {
  print("Job Usr #", id)
  lock.doWithLock {
    let x = mysql.query(statement: "select User from user")
    guard x, let y = mysql.storeResults(),
      let row = y.next() else {
       print(mysql.errorMessage())
       return
   }
    print(row[0] ?? "User() FAILED")
    y.close()
    jobs -= 1
  }
}

_ = mysql.setOption(.MYSQL_SET_CHARSET_NAME, "utf8mb4")
guard mysql.connect(host: "127.0.0.1", user: "root", password: "your pass", db: "mysql") else {
  print(mysql.errorMessage())
  exit(0)
}

jobs = 10
for id in 0 ..< 5 {
  Threading.dispatch {
    now(id)
  }
  Threading.dispatch {
    user(id)
  }
}

while jobs > 0 {
  sleep(1)
}

By using Threading.Lock(), you can easily queue all queries into one connection and maximize the efficiency - fast, atomic, ordered, shared and singleton.

If insisted, you can create a limited number of connections to an array, and each connection keeps one lock to make a balance, let say, a threaded connection pool, so I believe this balanced solution can be established without any special module on Perfect itself:

struct MySQLPool {
  let mysql = MySQL()
  let lock = Threading.Lock()
 }
var pool: [MySQLPool] = []
for _ in 0 ..< 32 {
  pool.append(YourNewConnection)
}