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Question

所以我处于学习swift的早期阶段，并且我试图创建一个简单的类来包装从给定的Web服务发送/检索数据的过程。我遇到的问题是，在发送请求或任何类型的响应后，没有任何内容正在打印到控制台。我真的很感激任何关于我做错的帮助或指导

import Foundation

class URLHelper : NSObject,NSURLConnectionDelegate,NSURLConnectionDataDelegate{

var data = NSMutableData()


func sendReq(){
    let urlPath: String = "http://localhost/web-service/action.php?callback=showUserDetails&uid=1"
    var url: NSURL = NSURL(string: urlPath)!
    var request: NSURLRequest = NSURLRequest(URL: url,cachePolicy: NSURLRequestCachePolicy.ReloadIgnoringLocalCacheData, timeoutInterval: 4)
    var connection: NSURLConnection = NSURLConnection(request: request, delegate: self, startImmediately: false)!

    connection.start()
}


func connection(connection: NSURLConnection!, didReceiveData data: NSData!){
    self.data.appendData(data)
}

func connection(connection: NSURLConnection, didFailWithError error: NSError) {
    println(error.description)
}

func connectionDidFinishLoading(connection: NSURLConnection!) {
    var err: NSError
    var jsonResult: NSDictionary = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.MutableContainers, error: nil) as NSDictionary
    println(jsonResult)
}


}


var req = URLHelper()   
req.sendReq()

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<?php



//Get the action to run the coorect request
if(isset($_GET['callback'])){
        $function = $_GET['callback'];
        call_user_func($function);
        //$function();
}else{
    echo "Error: No valid callback supplied to request";
}


function showUserDetails(){ 
    $conn = mysqli_connect("localhost", "root", "root", "service_db") or die("Error " . mysqli_error($conn)); 
    $userid = $_GET['uid'];
    $results = mysqli_fetch_assoc(mysqli_query($conn,"SELECT * FROM user WHERE id = $userid")); 
    mysqli_close($conn);

    echo json_encode($results);

}


if($db->connect_errno > 0){
  die('Unable to connect to database [' . $db->connect_error . ']');
 }

?>

返回的json如下{"id":"1","username":"tom","email":"tom_smith@gmail.com"}

Answer 1

对于发布，您无需提出请求，只需从网址获取内容：

        var strResult:NSString

        let strURL = "http://localhost/web-service/action.php?callback=showUserDetails&uid=1"

        var dataURL = NSData(contentsOfURL: NSURL(string: strURL)!);

        if let d = dataURL
        {
            strResult = NSString(data: d, encoding: NSUTF8StringEncoding)!
            println(strResult)

        }

在这里，我只需转到网址，获取内容并将其存储在数据对象中，然后将其转换为字符串并打印即可。您还可以将JSON字符串（strResult）解码为JSON对象。

希望有所帮助：）

Answer 2

您可以使用具有completionHandler的此方法从Web服务中获取结果

 func postAsync(backendMethodName:String ,body: [[String:String]], completionHandler: (resultnig:String) -> Void)
{

    let session: NSURLSession = NSURLSession.sharedSession()
    let urlPath: String = "\(sessionclass.connectionString)/\(backendMethodName)"

    let request = NSMutableURLRequest(URL: NSURL(string:urlPath)!)

    request.HTTPMethod = "POST"
    request.timeoutInterval=10
    request.addValue("application/json", forHTTPHeaderField: "Content-Type")
    request.addValue("application/json", forHTTPHeaderField: "Accept")
    request.HTTPBody = try! NSJSONSerialization.dataWithJSONObject(body, options: [])

    let task = session.dataTaskWithRequest(request) { data, response, error in
        guard data != nil else
        {
            print("no data found: \(error)")
            return
        }

        let strData = NSString(data: data!, encoding: NSUTF8StringEncoding)
        completionHandler(resultnig: strData as! String)


    }

    task.resume()

}

它采用后端方法Name（在Web服务中）和你的jsonData

你可以这样称呼它：

 self.postAsync("checkConnection", body: self.alldictionariesConn, completionHandler: { (resultnig) in
       print(resulting) })//resulting = result from your web service

Swift发送请求没有响应

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2 个答案: