Question

这是我的php api代码：

header('Content-Type: application/json');
$return_arr = array();  
$username=mysqli_real_escape_string($link,$_POST["name"]);
$email=mysqli_real_escape_string($link,$_POST["email"]);
$mobile=mysqli_real_escape_string($link,$_POST["mobile"]);
$password=mysqli_real_escape_string($link,$_POST["password"]);

$result=mysqli_query($link,"insert into users(name,email,mob,password) values('$username','$email','$mobile','$password')");    
if(mysqli_num_rows($result))
{
$row_array['status']=true;
array_push($return_arr,$row_array);
}
else 
{
$row_array['status']=false;
array_push($return_arr,$row_array);
}
echo json_encode($return_arr);
mysqli_close($link);

Nil条目正在插入数据库。实际问题在哪里，我测试了所有但是无法获得。

此代码无效，此代码中的url编码方式如何：

let signUpUrl = URL(string: "myURL/api_register.php")
var request = URLRequest(url: signUpUrl!)
request.httpMethod = "POST"
request.addValue("application/json", forHTTPHeaderField: "Content-Type")
request.addValue("application/json", forHTTPHeaderField: "Accept")


let postString : Parameters = ["name": "com", "email": "raj123@v", "mobile": "123", "password": "123"]
print("\(postString)")
do {
    request.httpBody = try JSONSerialization.data(withJSONObject: postString, options: .prettyPrinted)
    //print(request.description)
} catch let error {
    print(error.localizedDescription)
    displayMessage(userMessage: "Something went wrong")
    return
}
 let task = URLSession.shared.dataTask(with: request) { (data: Data?, response: URLResponse?, error: Error?) in



    if error != nil {
        self.displayMessage(userMessage: "Count perform this request, Please try again later")
        print("Error desription  \(String(describing: error))")
        return
    }
    guard let data = data else{
        print("no data return")
        return
    }
    //Let's convert response send from the server side to NSDictionary
    do {
        let json = try JSONSerialization.jsonObject(with: data) as? [String: Any]
        print("json is..\(String(describing: json))")

        // Now we can access value of User ID by its key
        if let printedJson = json {
            print("print... \(printedJson)")
            let userId = printedJson["success"] as? Bool
            print("user id is \(String(describing: userId))")
            if !userId! {

                //self.displayMessage(userMessage: "Cound not perform this request, Please try again.")
                //return
                print("no.")
            } else {
                //Redirect to Home Screen----------------------------
                print("yes")

                // let appdelegate = UIApplication.shared.delegate
                //  appdelegate?.window??.rootViewController = homepage
            }
        } else {
            self.displayMessage(userMessage: "Cound not perform this request, Please try again...")
            return
        }


    } catch {

        self.displayMessage(userMessage: "Cound not perform this request, Please try again.")
        print("error\(error)")
        return

    }


}
task.resume()

当我使用带有URLEncoding.default方法的alarmofire.request时，它的工作却使用了它。这段代码中缺少一些东西。

数据没有插入swift 4中的数据库（nil数据插入带有空白条目的数据库中）

0 个答案: