线程1：在使用Alamofire进行登录时发出SIGABRT信号。

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线程1：在使用Alamofire进行登录时发出SIGABRT信号。

我试图用Alamofire制作一个登录按钮，但我得到了标题中提到的错误，我的代码有什么问题吗？谢谢。

@IBAction func buttonLogin(_ sender: UIButton) {

//获取用户名和密码

let parameters: Parameters=[

"usuario":textFieldUserName.text!,

"password":textFieldPassword.text!

]

//发起POST请求

Alamofire.request("一个JSON页面的URL", method: .post, parameters: parameters).responseString { response in

//打印响应

print(response)

//从服务器获取JSON值

if let result = response.result.value {

let jsonData = result as NSObject

//从响应中获取用户

let user = jsonData.value(forKey: "user") as! NSArray

//获取用户值

let userId = user.value(forKey: "nombre") as! String

let userName = user.value(forKey: "usuario") as! String

let userPassword = user.value(forKey: "password") as! String

//将用户值保存到默认值

self.defaultValues.set(userId, forKey: "nombre")

self.defaultValues.set(userName, forKey: "usuario")

self.defaultValues.set(userPassword, forKey: "password")

//切换屏幕

let profileViewController = self.storyboard?.instantiateViewController(withIdentifier: "ProfileViewcontroller") as! ProfileViewController

self.navigationController?.pushViewController(profileViewController, animated: true)

self.dismiss(animated: false, completion: nil)

}else{

//无效凭证时的错误消息

self.labelMessage.text = "用户名或密码无效"

}

}

}

打印错误为：

*** Terminating app due to uncaught exception 'NSUnknownKeyException', reason: '[ valueForUndefinedKey:]: this class is not key value coding-compliant for the key user.'
*** First throw call stack:
(0x18542a364 0x184670528 0x18542a02c 0x185dde434 0x185d24e20 0x1003e61a4 0x100abb708 0x100ab797c 0x100a76d38 0x10188d2cc 0x10188d28c 0x101891ea0 0x1853d2544 0x1853d0120 0x1852efe58 0x18719cf84 0x18e96f67c 0x1003e9500 0x184e0c56c)
libc++abi.dylib: terminating with uncaught exception of type NSException

发生在：

//从响应中获取用户

let user = jsonData.value(forKey: "user") as! NSArray

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