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I want to POST Data in a Textfield via HTTPS to a Webserver using Swift and PHP and tried this Example POST data to a PHP method from Swift, but it is not compatible with PHP7.

Now this is for PHP7 and Swift 4 but i got an empty entry in the MySql Database.

I think thats the PHP-File...

Use Xcode 10.1 Any Solution?

Swift:

import UIKit class MessageViewController: UIViewController { @IBOutlet weak var nachricht: UITextField! @IBAction func submit(_ sender: Any) { let url = NSURL(string: "http://localhost.com") // localhost MAMP - change to point to your database server var request = URLRequest(url: url! as URL) request.httpMethod = "POST" var dataString = "secretWord=???" // starting POST string with a secretWord // the POST string has entries separated by & dataString = dataString + "&nachricht=\(nachricht.text!)" // add items as name and value // convert the post string to utf8 format let dataD = dataString.data(using: .utf8) // convert to utf8 string do { // the upload task, uploadJob, is defined here let uploadJob = URLSession.shared.uploadTask(with: request, from: dataD) { data, response, error in if error != nil { // display an alert if there is an error inside the DispatchQueue.main.async DispatchQueue.main.async { let alert = UIAlertController(title: "Upload Didn't Work?", message: "Looks like the connection to the server didn't work. Do you have Internet access?", preferredStyle: .alert) alert.addAction(UIAlertAction(title: "OK", style: .cancel, handler: nil)) self.present(alert, animated: true, completion: nil) } } else { if let unwrappedData = data { let returnedData = NSString(data: unwrappedData, encoding: String.Encoding.utf8.rawValue) // Response from web server hosting the database if returnedData == "1" { // insert into database worked // display an alert if no error and database insert worked (return = 1) inside the DispatchQueue.main.async DispatchQueue.main.async { let alert = UIAlertController(title: "Upload OK?", message: "Looks like the upload and insert into the database worked.", preferredStyle: .alert) alert.addAction(UIAlertAction(title: "OK", style: .cancel, handler: nil)) self.present(alert, animated: true, completion: nil) } } else { // display an alert if an error and database insert didn't worked (return != 1) inside the DispatchQueue.main.async DispatchQueue.main.async { let alert = UIAlertController(title: "Upload Didn't Work", message: "Looks like the insert into the database did not worked.", preferredStyle: .alert) alert.addAction(UIAlertAction(title: "OK", style: .cancel, handler: nil)) self.present(alert, animated: true, completion: nil) } } } } } uploadJob.resume() } } }

PHP File:

<?php $secret = $_POST["secretWord"]; if ("???" != $secret) exit; // note the same secret as the app - could be let out if this check is not required. secretWord is not entered by the user and is used to prevent unauthorized access to the database $nachricht = $_POST['nachricht']; // POST items should be checked for bad information before being added to the database. // Create connection $mysqli=mysqli_connect("localhost","db","db_pass","db_usr"); // localhost, user name, user password, database name // Check connection if (mysqli_connect_errno()) { echo " Failed to connect to MySQL: " . mysqli_connect_error(); } $query = "insert into `db` (nachricht) value ('".$nachricht."')"; $result = mysqli_query($mysqli,$query); echo $result; // sends 1 if insert worked ?>
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  • The PHP version on your server should be entirely irrelevant to your problem. What's the exact problem you are having? Commented Nov 28, 2018 at 20:24
  • The Problem is, that i got an empty entry in the Database. The written Text in the Swift Textfield will not be transmitted to the Database. Commented Nov 29, 2018 at 8:07
  • I can't understand the "$query" part. What are you trying to do exactly? Commented Dec 3, 2018 at 7:16
  • Oh, sorry. I have updated the code. I want to pass the text (nachricht) from the app, via the php file of a database Commented Dec 4, 2018 at 12:03

3 Answers 3

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I do not have much info on the PHP part of your problem. But i would recommend you try using "https://github.com/Alamofire/Alamofire" for all your networking needs as it supports Swift 4 with PHP7. It also reduces the number of lines of code you need to write.

So with Alamofire your POST request would look something like this.

import UIKit import Alamofire class MessageViewController: UIViewController { @IBOutlet weak var nachricht: UITextField! @IBAction func submit(_ sender: Any) { let body: Parameters = [ "secretWord":"???", "nachricht":"\(nachricht.text!)", ] Alamofire.request("http://localhost.com", method: .post, parameters: body, encoding: JSONEncoding.default).responseJSON { response in print("Your response from server is: \(response)") } } }

Hope this solves the Swift part of your problem.

EDIT: As requested here is an update. I have implemented your request without using a networking manager (Alamofire)

import UIKit class MessageViewController: UIViewController { @IBOutlet weak var nachricht: UITextField! @IBAction func submit(_ sender: Any) { // MARK: Preparing json data let parameters: [String: Any] = [ "secretWord":"???", "nachricht":"\(nachricht.text!)", ] //Serialise Parameters to JSON data. let jsonData = try? JSONSerialization.data(withJSONObject: parameters) // create post request let url = URL(string: "http://localhost.com")! var request = URLRequest(url: url) request.httpMethod = "POST" // insert json data to the request request.httpBody = jsonData let task = URLSession.shared.dataTask(with: request) { data, response, error in print("Session Response: \(String(describing: response!))") let httpResponse = response as? HTTPURLResponse let statusCode = httpResponse?.statusCode if statusCode == 404 { DispatchQueue.main.async { // create the alert let alert = UIAlertController(title: "Server not available", message: "Try again later.", preferredStyle: UIAlertController.Style.alert) // add an action (button) alert.addAction(UIAlertAction(title: "OK", style: UIAlertAction.Style.default, handler: nil)) // show the alert self.present(alert, animated: true, completion: nil) } }else{ guard let data = data, error == nil else { print(error?.localizedDescription ?? "No data") return } let responseJSON = try? JSONSerialization.jsonObject(with: data, options: []) if let responseJSON = responseJSON as? [String: Any] { print("Your response from server is: \(responseJSON)") } } } task.resume() } }
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6 Comments

Thanks for the help, but I would like to have my own script, server and so one. I've used firebase before.
@Humfeld I have edited my answer and implemented your request without using any networking manager. Please check.
You are nice, but it did not work. The code works without problems. Still no entry in the Database. This is from Xcode:
@Humfeld According to the Session Response you seem to be getting a 200 Status Code. Which means the API call was successful but it did not send any response. So your swift code is perfectly fine. You need to look in to the backend part of you code. The PHP code is not accepting the POST data.
Hmm, So the bug is in the php file? Is there another solution
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This is from Xcode: [MC] Reading from private effective user settings. Session Response: { URL: https://localhost.com/api.php } { Status Code: 200, Headers { Connection = ( "Keep-Alive" ); "Content-Encoding" = ( gzip ); "Content-Length" = ( 21 ); "Content-Type" = ( "text/html; charset=UTF-8" ); Date = ( "Sun, 02 Dec 2018 10:06:45 GMT" ); "Keep-Alive" = ( "timeout=5, max=100" ); Server = ( Apache ); "Strict-Transport-Security" = ( "max-age=31556926" ); Vary = ( "Accept-Encoding" ); } } 2018-12-02 11:10:36.854962+0100 Localhost[2289:36190] [BoringSSL] nw_protocol_boringssl_get_output_frames(1301) [C1.1:2][0x7f8f76f215b0] get output frames failed, state 8196

There is no entry from the Text field transmitted in the MySql Database.

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I had the same problem. My solution was to use import MultipartForm and define the form.

let form = MultipartForm(parts: [ MultipartForm.Part(name: "id", value: incidente!.id), MultipartForm.Part(name: "mensaje", value: txtDescripcion.text!), MultipartForm.Part(name: "direccion", value: txtDireccion.text!), MultipartForm.Part(name: "nombre", value: txtNombre.text!), MultipartForm.Part(name: "telefono", value: txtTelefono.text!), MultipartForm.Part(name: "email", value: txtEmail.text!), MultipartForm.Part(name: "latitud", value: StrLatitud), MultipartForm.Part(name: "longitud", value: StrLongitud), MultipartForm.Part(name: "token", value: "44fdcv8jf3"), MultipartForm.Part(name: "imagen64", value: mImagenStr) ])

Then save the image with

file_put_contents($ruta,base64_decode($_POST['imagen64']));
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