發送圖像PHP迅速

Question

我試圖與SWIFT 我的本地PHP服務器上發送的圖像，我不明白爲什麼我有一個錯誤消息（致命錯誤：意外發現零而展開的可選值）

這裏是我的代碼：

var image = info[UIImagePickerControllerOriginalImage] as UIImage 
var imageData = UIImageJPEGRepresentation(image, 90) 

var request = NSMutableURLRequest(URL: NSURL(string:"http://http:192.168.1.73:8888/upload.php")!) 
var session = NSURLSession.sharedSession() 

request.HTTPMethod = "POST" 

var boundary = NSString(format: "---------------------------14737809831466499882746641449") 
var contentType = NSString(format: "multipart/form-data; boundary=%@",boundary) 

request.addValue(contentType, forHTTPHeaderField: "Content-Type") 

var body = NSMutableData.alloc() 


// Image 
body.appendData(NSString(format: "\r\n--%@\r\n", boundary).dataUsingEncoding(NSUTF8StringEncoding)!) 
body.appendData(NSString(format:"Content-Disposition: form-data; name=\"htdocs\"; filename=\".jpg\"\r\n").dataUsingEncoding(NSUTF8StringEncoding)!) 
body.appendData(NSString(format: "Content-Type: application/octet-stream\r\n\r\n").dataUsingEncoding(NSUTF8StringEncoding)!) 
body.appendData(imageData) 
body.appendData(NSString(format: "\r\n--%@\r\n", boundary).dataUsingEncoding(NSUTF8StringEncoding)!) 



request.HTTPBody = body 


var returnData = NSURLConnection.sendSynchronousRequest(request, returningResponse: nil, error: nil) 
println(returnData) 
var returnString = NSString(data: returnData!, encoding: NSUTF8StringEncoding) 

println("returnString \(returnString)") 

Answer 1

你不能強迫一個Optional值的解纏，因爲它可能是nil，而不是一個真正的價值。

走這條線，例如：

var request = NSMutableURLRequest(URL: NSURL(string:"http://http:192.168.1.73:8888/upload.php")!) 

如果NSURL(string:"http://http:192.168.1.73:8888/upload.php")回報nil，你會打電話NSMutableURLRequest(URL: nil)這是無效的。

您需要檢查是否展開的值是有效的：

if let url = NSURL(string:"http://http:192.168.1.73:8888/upload.php") { 
    var request = NSMutableURLRequest(URL: url) 
}