从urlsession返回数据并保存属性变量

我尝试使用 urlsession.shared.datatask 从服务器获取一些数据。它可以正常工作，但是我无法像类变量那样保存结果。许多答案建议使用完成处理程序，但这对我的任务无济于事。

这是我的测试代码：

class PostForData {
func forData(completion:  @escaping (String) -> ()) {
    if let url = URL(string: "http://odnakrov.info/MyWebService/api/test.php") {
        var request = URLRequest(url: url)
        request.httpMethod = "POST"
        let postString : String = "json={"Ivan Bolgov":"050-062-0769"}"
        print(postString)
        request.httpBody = postString.data(using: .utf8)
        let task = URLSession.shared.dataTask(with: request) {
            data, response, error in
            let json = String(data: data!, encoding: String.Encoding.utf8)!
                completion(json)
        }
        task.resume()
    }
}
}
class ViewController: UIViewController {
var str:String?
override func viewDidLoad() {
    super.viewDidLoad()
    let pfd = PostForData()
    pfd.forData { jsonString in
        print(jsonString)
        DispatchQueue.main.async {
            self.str = jsonString
        }
    }
    print(str ?? "not init yet")
}
}

此闭合是 @escaping(即，稍后调用异步(，因此您必须将其放入闭合中：

class ViewController: UIViewController {
    @IBOutlet weak var label: UILabel!
    var str: String?
    override func viewDidLoad() {
        super.viewDidLoad()
        let pfd = PostForData()
        pfd.performRequest { jsonString, error in
            guard let jsonString = jsonString, error == nil else {
                print(error ?? "Unknown error")
                return
            }
            // use jsonString inside this closure ...
            DispatchQueue.main.async {
                self.str = jsonString
                self.label.text = jsonString
            }
        }
        // ... but not after it, because the above runs asynchronously (i.e. later)
    }
}

注意，我更改了您的闭合，以返回String?和Error?，以便视图控制器可以知道是否发生了错误(如果它在乎，它可以看到发生了什么样的错误(。

。

注意，我将您的forData重命名为performRequest。通常，您会使用比这更有意义的名称，但是方法名称(在Swift 3和更高版本中(通常应包含一个动词，该动词指示正在完成的操作。

class PostForData {
    func performRequest(completion:  @escaping (String?, Error?) -> Void) {
        // don't try to build JSON manually; use `JSONSerialization` or `JSONEncoder` to build it
        let dictionary = [
            "name": "Ivan Bolgov",
            "ss": "050-062-0769"
        ]
        let jsonData = try! JSONEncoder().encode(dictionary)
        // It's a bit weird to incorporate JSON in `x-www-form-urlencoded` request, but OK, I'll do that.
        // But make sure to percent escape it.
        let jsonString = String(data: jsonData, encoding: .utf8)!
            .addingPercentEncoding(withAllowedCharacters: .urlQueryValueAllowed)!
        let body = "json=" + jsonString
        let url = URL(string: "http://odnakrov.info/MyWebService/api/test.php")!
        var request = URLRequest(url: url)
        request.httpMethod = "POST"
        request.httpBody = body.data(using: .utf8)
        // It's not required, but it's good practice to set `Content-Type` (to specify what you're sending)
        // and `Accept` (to specify what you're expecting) headers.
        request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
        request.setValue("application/json", forHTTPHeaderField: "Accept")
        // now perform the prepared request
        let task = URLSession.shared.dataTask(with: request) { data, _, error in
            guard let data = data, error == nil else {
                completion(nil, error)
                return
            }
            let responseString = String(data: data, encoding: .utf8)
            completion(responseString, nil)
        }
        task.resume()
    }
}

该例程也有一些修改，特别是：

在处理服务器响应时，永远不要使用!强制解析。您无法控制请求是成功还是失败，并且强制解开操作员会崩溃您的应用程序。您应该通过guard let或if let模式优雅地解开这些选项。
使用...是JSON字符串，使用json=...模式非常不寻常。可以从此推断出您正在准备application/x-www-form-urlencoded请求，然后使用$_POST或$_REQUEST来获取与json密钥相关的值。通常，您要么要执行真正的JSON请求，要么您要执行application/x-www-form-urlencoded请求，但两者都不是。但是，在一个请求中同时做的是将客户端和服务器代码中的工作量增加一倍。以上代码遵循您的原始代码段中的模式，但我建议使用一个或另一个，但不要同时使用。
就我个人而言，我不会返回JSON字符串。我建议它实际上执行JSON的解析。但是，我再次离开了它，因为它在您的代码段中。
我注意到您以"Ivan Bolgov": "050-062-0769"的形式使用了JSON。我建议不要使用"值"作为JSON的钥匙。键应为有优势定义的常数。因此，例如，上面我使用了"name": "Ivan Bolgov"和"ss": "050-062-0769"，服务器知道在其中寻找称为name和ss的密钥。在这里做任何您想做的事，但是您的原始JSON请求似乎将密钥(通常是事先知道的(和值(与这些密钥相关联的值(。

如果您要执行x-www-form-urlencoded请求，则必须像我上面一样编码所提供的值。值得注意的是，在这些请求中不允许像空间字符这样的字符，因此您必须对其进行评估。不用说，如果您做了适当的JSON请求，则不需要这种愚蠢。

但请注意，当编码百分比时，不要试图使用默认的.urlQueryAllowed字符集，因为它将允许某些字符通过UneScaped。因此，我定义了.urlQueryValueAllowed，该CC_24从.urlQueryAllowed字符集中删除某些字符(改编自Alamofire中使用的模式(：

extension CharacterSet {
    /// Returns the character set for characters allowed in the individual parameters within a query URL component.
    ///
    /// The query component of a URL is the component immediately following a question mark (?).
    /// For example, in the URL `http://www.example.com/index.php?key1=value1#jumpLink`, the query
    /// component is `key1=value1`. The individual parameters of that query would be the key `key1`
    /// and its associated value `value1`.
    ///
    /// According to RFC 3986, the set of unreserved characters includes
    ///
    /// `ALPHA / DIGIT / "-" / "." / "_" / "~"`
    ///
    /// In section 3.4 of the RFC, it further recommends adding `/` and `?` to the list of unescaped characters
    /// for the sake of compatibility with some erroneous implementations, so this routine also allows those
    /// to pass unescaped.
    static var urlQueryValueAllowed: CharacterSet = {
        let generalDelimitersToEncode = ":#[]@"    // does not include "?" or "/" due to RFC 3986 - Section 3.4
        let subDelimitersToEncode = "!$&'()*+,;="
        var allowed = CharacterSet.urlQueryAllowed
        allowed.remove(charactersIn: generalDelimitersToEncode + subDelimitersToEncode)
        return allowed
    }()
}

我建议更改您的PHP以接受JSON请求，例如：

<?php
    // read the raw post data
    $handle = fopen("php://input", "rb");
    $raw_post_data = '';
    while (!feof($handle)) {
        $raw_post_data .= fread($handle, 8192);
    }
    fclose($handle);
    // decode the JSON into an associative array
    $request = json_decode($raw_post_data, true);
    // you can now access the associative array how ever you want
    if ($request['foo'] == 'bar') {
        $response['success'] = true;
        $response['value']   = 'baz';
    } else {
        $response['success'] = false;
    }
    // I don't know what else you might want to do with `$request`, so I'll just throw
    // the whole request as a value in my response with the key of `request`:
    $raw_response = json_encode($response);
    // specify headers
    header("Content-Type: application/json");
    header("Content-Length: " . strlen($raw_response));
    // output response
    echo $raw_response;
?>

然后，您可以简化请求的构建，从而消除了我们与x-www-form-urlencoded请求有关的所有代码的需求：

class PostForData {
    func performRequest(completion:  @escaping (String?, Error?) -> Void) {
        // Build the json body
        let dictionary = [
            "name": "Ivan Bolgov",
            "ss": "050-062-0769"
        ]
        let data = try! JSONEncoder().encode(dictionary)
        // build the request
        let url = URL(string: "http://odnakrov.info/MyWebService/api/test.php")!
        var request = URLRequest(url: url)
        request.httpMethod = "POST"
        request.httpBody = data
        // It's not required, but it's good practice to set `Content-Type` (to specify what you're sending)
        // and `Accept` (to specify what you're expecting) headers.
        request.setValue("application/json", forHTTPHeaderField: "Content-Type")
        request.setValue("application/json", forHTTPHeaderField: "Accept")
        // now perform the prepared request
        let task = URLSession.shared.dataTask(with: request) { data, _, error in
            guard let data = data, error == nil else {
                completion(nil, error)
                return
            }
            let responseString = String(data: data, encoding: .utf8)
            completion(responseString, nil)
        }
        task.resume()
    }
}

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