[英]Swift4 - Error Domain=NSCocoaErrorDomain Code=4865
func DoLogin(_ email:String, _ password:String)
{
struct user : Decodable {
let userid: Int
let sfname: String
let slname: String
let email: String
let sid: Int
}
let url = URL(string: ".....")!
var request = URLRequest(url: url)
request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
request.httpMethod = "POST"
let postString = "email=" + email + "&password=" + password + ""
request.httpBody = postString.data(using: .utf8)
let task = URLSession.shared.dataTask(with: request) { data, response, error in
guard let data = data, error == nil else { // check for fundamental networking error
print(error!)
return
}
if let httpStatus = response as? HTTPURLResponse, httpStatus.statusCode != 200 { // check for http errors
print("statusCode should be 200, but is \(httpStatus.statusCode)")
print(response!)
}
let responseString = String(data: data, encoding: .utf8)
print(responseString!)
do {
let myStruct = try JSONDecoder().decode(user.self, from: data)
print(myStruct)
} catch let error as NSError {
print(error)
}
}
task.resume()
}
因此,目标是将JSON响应保存到“用户”类中,以便我可以使用变量将数据插入sql数据库。 我目前遇到的问题是错误消息...
“错误域= NSCocoaErrorDomain代码= 4865”没有与键userid(“ userid”)关联的值。“ UserInfo = {NSCodingPath =(),NSDebugDescription =没有与键userid(” userid“)关联的值。}”
我认为问题在于HTTP响应以数组形式返回数据,该数据随后无法被解码(下面列出的HTTP响应是我一直用于测试目的的responseString)
{"user":{"userid":2,"sfname":"John","slname":"Doe","email":"john@doe.com","sid":123456}}
这是用于返回数据的PHP。
public function getUserByEmail($email)
{
$stmt = $this->conn->prepare("SELECT userid, sfname, slname, email, sid FROM students WHERE email = ?");
$stmt->bind_param("s", $email);
$stmt->execute();
$stmt->bind_result($userid, $sfname, $slname, $email, $sid);
$stmt->fetch();
$user = array();
$user['userid'] = $userid;
$user['sfname'] = $sfname;
$user['slname'] = $slname;
$user['email'] = $email;
$user['sid'] = $sid;
return $user;
}
提前致谢
正如您已经提到的那样,JSON的结构与user
结构的结构不匹配。
您可以做两件事:尝试弄清为什么将返回的JSON包装在另一个JSON对象中,或者创建一个与接收的JSON结构匹配的包装器结构。
第二种方法应产生如下结果:
struct UserWrapper: Decodable {
let user:user
}
然后,当您从JSON创建用户时,只需
let wrapper = try JSONDecoder().decode(UserWrapper.self, from: data)
myStruct = wrapper.user
顺便说一句:我建议您阅读Swift风格指南。 按照约定,函数名称应以小写字母开头,而类/结构名称应以大写字母开头。 在函数中也命名为参数的文档是很酷的事情。
就像是
func doLogin(userWithMail email: String, andPassword password: String) {...}
// ... other stuff
// call
doLogin(userWithMail: "test@test.com", andPassword: "1234567")
可能更具可读性。 以防万一您将来必须共享您的代码库。 ; d
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