如何在游乐场中运行异步回调

Question

// Playground - noun: a place where people can play

import Cocoa
import XCPlayground

let url = NSURL(string: "http://stackoverflow.com")
let request = NSURLRequest(URL: url)

NSURLConnection.sendAsynchronousRequest(request, queue:NSOperationQueue.currentQueue() {
response, maybeData, error in

    // This block never gets called?
    if let data = maybeData {
        let contents = NSString(data:data, encoding:NSUTF8StringEncoding)
        println(contents)
    } else {
        println(error.localizedDescription)
    }
}

Answer 1

虽然您可以手动运行一个运行循环（或者对于不需要运行循环的异步代码，使用其他等待方法，如调度信号量），但是我们在playgrounds中提供的等待异步工作的“内置”方式是导入xcplayground框架并设置xcplaygroundPage.currentPage.needsinDefiniteExecution=true。如果已经设置了此属性，那么当顶级游乐场源完成时，我们将继续旋转主运行循环，而不是停止游乐场，这样异步代码就有机会运行。我们最终会在超时后终止游乐场，该超时默认为30秒，但如果您打开助理编辑器并显示时间线助理，可以配置该超时；超时在右下角。

例如，在Swift 3中（使用URLSession而不是NSURLConnection):

import UIKit
import PlaygroundSupport

let url = URL(string: "http://stackoverflow.com")!

URLSession.shared.dataTask(with: url) { data, response, error in
    guard let data = data, error == nil else {
        print(error ?? "Unknown error")
        return
    }

    let contents = String(data: data, encoding: .utf8)
    print(contents!)
}.resume()

PlaygroundPage.current.needsIndefiniteExecution = true

或者在Swift 2中：

import UIKit
import XCPlayground

let url = NSURL(string: "http://stackoverflow.com")
let request = NSURLRequest(URL: url!)

NSURLConnection.sendAsynchronousRequest(request, queue: NSOperationQueue.currentQueue()) { response, maybeData, error in
    if let data = maybeData {
        let contents = NSString(data:data, encoding:NSUTF8StringEncoding)
        println(contents)
    } else {
        println(error.localizedDescription)
    }
}

XCPlaygroundPage.currentPage.needsIndefiniteExecution = true