Question

在这里，我尝试使用手机号码注册应用程序，但我无法以正确的格式进行，我在Json解析注册应用程序时出错。

在这里，我给出了我尝试过的代码，

var request = NSMutableURLRequest(URL: NSURL(string: "http://app.mycompany.in/gcm/test_slim.php/register")!, cachePolicy: NSURLRequestCachePolicy.ReloadIgnoringLocalCacheData, timeoutInterval: 5)

var response: NSURLResponse?
var error: NSError?
var reponseError: NSError?

var urlData: NSData? = NSURLConnection.sendSynchronousRequest(request, returningResponse:&response, error:&reponseError)

// create some JSON data and configure the request

let jsonString = "json=[{\"gsm_number\":\(Mobile),\"name\":\(Name),\"email\":\(Email),\"status\":\(Status),\"ver_code\":,\"profile_picture\":,\"device_id\":,\"gcm\":,\"is_register\":,\"thumb_image\":,\"user_baground_img\":}]"

request.HTTPBody = jsonString.dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: true)
request.HTTPMethod = "POST"
request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")

// send the request
NSURLConnection.sendSynchronousRequest(request, returningResponse: &response, error: &error)

// look at the response
if let httpResponse = response as? NSHTTPURLResponse {
    println("HTTP response: \(httpResponse.statusCode)")
    println("jsonString: \(jsonString)")
    var responseData:NSString  = NSString(data:urlData!, encoding:NSUTF8StringEncoding)!

    let VerificationcodeViewController = self.storyboard?.instantiateViewControllerWithIdentifier("verificationcodeViewController") as UIViewController

    self.navigationController?.pushViewController(VerificationcodeViewController, animated: true)
} else {
    println("No HTTP response")
    var alertView:UIAlertView = UIAlertView()
    alertView.title = "Error!"
    alertView.message = "Error. & Some Problem was Found"
    alertView.delegate = self
    alertView.addButtonWithTitle("OK")
    alertView.show()
}

Answer 1

你不能有空钥匙，试试这个：

let parameters = [
    "gsm_number" : Mobile,
    "name" : Name,
    "email" : Email,
    "status" : Status,
]

let jsonData = NSJSONSerialization.dataWithJSONObject(parameters, options: NSJSONWritingOptions.allZeros, error: nil)

let jsonString = "json=\(NSString(data: jsonData!, encoding: NSUTF8StringEncoding)!)"

Answer 2

您的jsonString不符合JSON语法。

您不能在json=开头，=不是JSON中允许的分隔符，请使用:
你需要用双引号包装字符串变量（并转义它们）
如果缺少值，您就无法拥有空键，您必须使用null

变量的有效字符串示例：

let jsonString = "[{\"gsm_number\":\(Mobile),\"name\":\"\(Name)\",\"email\":\"\(Email)\",\"status\":\(Status),\"ver_code\":null}]"

或者使用＆＃39; json＆＃39;一开始就像你一样关键：

let jsonString = "{\"json\":[{\"gsm_number\":\(Mobile),\"name\":\"\(Name)\",\"email\":\"\(Email)\",\"status\":\(Status),\"ver_code\":null}]}"

将其放入Playground并在＆＃34; View＆＃34;中显示助理编辑器。菜单，它会帮助您理解：

let Mobile = 42
let Name = "James"
let Email = "test@test.com"
let Status = 200
let jsonString = "[{\"gsm_number\":\(Mobile),\"name\":\"\(Name)\",\"email\":\"\(Email)\",\"status\":\(Status),\"ver_code\":null}]"
let data = jsonString.dataUsingEncoding(NSUTF8StringEncoding, allowLossyConversion: false)

Swift 1

var err: NSError?
let json = NSJSONSerialization.JSONObjectWithData(data!, options: NSJSONReadingOptions.allZeros, error: &err) as? [[String:AnyObject]]
if err != nil {
    println(err)
} else {
    println(json![0]["status"]!)
}

Swift 2

do {
    if let data = data,
      let json = try NSJSONSerialization.JSONObjectWithData(data, options: []) as? [[String:AnyObject]] {
        if let status = json[0]["status"] as? Int {
            print(status)
        }
    }
} catch let error as NSError {
    print(error.localizedDescription)
}

在iOS swift Json解析如何在post方法

2 个答案: