ios - 从 URLSession 返回数据并保存在属性变量中
<p><p>我尝试使用 <strong><em>URLSession.shared.dataTask</em></strong> 从服务器获取一些数据。
它工作正常，但我不能像类变量一样保存结果。
许多答案建议使用 <strong><em>completion Handler</em></strong>，但这对我的任务没有帮助。 </p>
<p>这是我的测试代码:</p>
<pre><code>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")
}
}
</code></pre></p>
<br><hr><h1><strong>Best Answer-推荐答案</ strong></h1><br>
<p><p>这个闭包是<code>@escaping</code>(即稍后异步调用)，所以你必须把它放在闭包里面:</p>
<pre><code>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)
}
}
</code></pre>
<p>注意，我将您的闭包更改为返回 <code>String?</code> 和 <code>Error?</code> 以便 ViewController 可以知道是否发生错误(如果它关心，它可以看到发生了什么样的错误)。</p>
<p>注意，我将您的 <code>forData</code> 重命名为 <code>performRequest</code>。通常，您会使用比这更有意义的名称，但方法名称(在 Swift 3 及更高版本中)通常应该包含一个指示正在执行的操作的动词。</p>
<pre><code>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()
}
}
</code></pre>
<p>对该例程也有一些修改，具体来说:</p>
<ol>
<li><p>永远不要在处理服务器响应时使用 <code>!</code> 强制解包。您无法控制请求是成功还是失败，强制解包操作符会使您的应用程序崩溃。您应该使用 <code>guard let</code> 或 <code>if let</code> 模式优雅地解开这些可选项。</p></li>
<li><p>使用 <code>json=...</code> 模式非常不寻常，其中 <code>...</code> 是 JSON 字符串。可以推断您正在准备 <code>application/x-www-form-urlencoded</code> 请求，并使用 <code>$_POST</code> 或 <code>$_REQUEST</code> 来获取与 <code>json</code> 键关联的值。通常您要么执行真正的 JSON 请求，要么执行 <code>application/x-www-form-urlencoded</code> 请求，但不能同时执行。但是在一个请求中完成这两项工作会使客户端和服务器代码的工作量增加一倍。上面的代码遵循原始代码片段中的模式，但我建议使用其中一个，但不要同时使用。</p></li>
<li><p>就我个人而言，我不会让 <code>performRequest</code> 返回 JSON 字符串。我建议它实际上执行 JSON 的解析。但是，我再次将其保留在您的代码片段中。</p></li>
<li><p>我注意到您以 <code>"Ivan Bolgov": "050-062-0769"</code> 的形式使用 JSON。我建议不要使用“值”作为 JSON 的键。键应该是有利地定义的常量。因此，例如，上面我使用了 <code>"name": "Ivan Bolgov"</code> 和 <code>"ss": "050-062-0769"</code>，服务器知道要在其中查找称为 <code>name</code> 和 <code>ss</code> 的键。在这里做任何你想做的事，但你最初的 JSON 请求似乎将键(通常是预先知道的)和值(与这些键关联的值)混为一谈。</p></li>
<li><p>如果您要执行 <code>x-www-form-urlencoded</code> 请求，您必须对提供的值进行百分比编码，就像我上面所说的那样。值得注意的是，这些类型的请求中不允许使用空格字符等字符，因此您必须对它们进行百分比编码。不用说，如果你做了一个正确的 JSON 请求，就不需要这些愚蠢的东西了。</p>
<p>但请注意，在进行百分比编码时，不要试图使用默认的 <code>.urlQueryAllowed</code> 字符集，因为它会允许某些字符不转义地传递。所以我定义了一个 <code>.urlQueryValueAllowed</code>，它从 <code>.urlQueryAllowed</code> 字符集中删除某些字符(改编自 Alamofire 中使用的模式):</p>
<pre><code>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
}()
}
</code></pre> </li>
</ol>
<hr/>
<p>我建议更改您的 PHP 以接受 JSON 请求，例如:</p>
<pre><code><?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;
?>
</code></pre>
<p>然后您可以简化请求的构建，消除我们对 <code>x-www-form-urlencoded</code> 请求所做的所有百分比编码的需要:</p>
<pre><code>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()
}
}
</code></pre></p>
<p style="font-size: 20px;">关于ios - 从 URLSession 返回数据并保存在属性变量中，我们在Stack Overflow上找到一个类似的问题：
<a href="https://stackoverflow.com/questions/48890209/" rel="noreferrer noopener nofollow" style="color: red;">
https://stackoverflow.com/questions/48890209/
</a>
</p>
页:
[1]