按下 UILabel 时分配变量值以打开 Apple Maps 的问题

Question

本质上，我正在解析 JSON 数据并将其分配给一个名为 addressPressNow 的变量，然后我将在用户点击 UILabel 时执行以下函数：

目标是打开 Apple Maps，提供它包含的变量值。

因为我正在为一个变量分配一个地址，它将包含空格例如：3981 试驾 Cupertino CA 95014

注意：变量的值被正确传递，因为当我在 func tapFunction 中执行 print(addressPressNow) 时它打印正确。

@objc
func tapFunction(sender:UITapGestureRecognizer) {

    let targetURL = NSURL(string: "http://maps.apple.com/?q=" + addressPressNow)!


    UIApplication.shared.openURL(targetURL as URL)

}

问题是我在将变量应用于字符串 URL 时遇到问题，出现以下错误：

Thread 1: Fatal error: Unexpectedly found nil while unwrapping an Optional value

以下是我为变量赋值的方式：

struct FacilityInfo: Decodable {
    let address: String

class infoViewController: UIViewController {

    var addressPressNow : String = ""

override func viewDidLoad() {
    super.viewDidLoad()



    let tap = UITapGestureRecognizer(target: self, action: #selector(infoViewController.tapFunction))
    addressInfo.isUserInteractionEnabled = true
    addressInfo.addGestureRecognizer(tap)


    let url = URL(string: "https://test/test/example”)!

    let task = URLSession.shared.dataTask(with: url) { data, response, error in

        // ensure there is no error for this HTTP response
        guard error == nil else {
            print ("error: \(error!)")
            return
        }

        // ensure there is data returned from this HTTP response
        guard let data = data else {
            print("No data")
            return
        }

        // Parse JSON into array of Car struct using JSONDecoder


        guard let cars = try? JSONDecoder().decode([FacilityInfo].self, from: data), let secondCar = cars.first
        else {
            print("Error: Couldn't decode data into cars array")
            return
        }
        DispatchQueue.main.async {
            self.addressPressNow = secondCar.facility_address
        }

    }

Answer 1

你说的是

NSURL(string: "http://maps.apple.com/?q=" + addressPressNow)!

注意最后的感叹号。这意味着 "if there's a problem, crash me"。如果你真的崩溃了，你几乎不能抱怨；这就是你要求做的。

基本上，如果可以避免，从不使用NSURL(string:)。要形成有效的 URL，请使用 URL 组件构建它。并由 valid 个组件组成。（不可能说 facility_address 是否是一个有效的 URL 查询，因为你没有显示它是什么。）

示例：

var comp = URLComponents()
comp.scheme = "https"
comp.host = "maps.apple.com"
comp.queryItems = [URLQueryItem(name: "q", value: "1 Infinite Loop, Cupertino, CA")]
if let url = comp.url {
    print(url) // https://maps.apple.com?q=1%20Infinite%20Loop,%20Cupertino,%20CA

}

这给了我们一个有效的 URL 实际有效。

Answer 2

"I am assigning an address to a variable it will contain spaces"

如果地址包含空格，则使用字符串创建 NSURL 将会崩溃。可以用addingPercentEncoding解决问题

if let encodedAddress = addressPressNow.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) {
    let targetURL = NSURL(string: "http://maps.apple.com/?q=" + encodedAddress)!
    UIApplication.shared.openURL(targetURL as URL)
}

并且不要使用 NSURL 和强制展开。像这样更新

if let encodedAddress = addressPressNow.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed),
    let targetURL = URL(string: "http://maps.apple.com/?q=" + encodedAddress) {
        UIApplication.shared.openURL(targetURL)
}

根据 matt 的建议使用 URLComponents

let addressPressNow = "3981 Test Drive Cupertino CA 95014"
var components = URLComponents(string: "http://maps.apple.com")
components?.queryItems = [URLQueryItem(name: "q", value: addressPressNow)]
print(components?.url)//http://maps.apple.com?q=3981%20Test%20Drive%20Cupertino%20CA%2095014
if let targetURL = components?.url {
    UIApplication.shared.open(targetURL, options: [:], completionHandler: nil)
}

按下 UILabel 时分配变量值以打开 Apple Maps 的问题

Issue assigning variable value to open Apple Maps when pressing a UILabel

ios

swift

url

apple-maps