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我有一個應用程序的VC,獲取所有需要的變量,並應通過POST請求將它們傳遞到一個PHP文件,在那裏它們被存儲併發送到數據庫。問題出現在數據庫中沒有設置變量(我相信連接已經完成)。這個php文件和一個Android應用程序一樣工作正常(所以變量使用Android應用程序很好地存儲)。快速發佈POST請求沒有什麼
如果您能給我一些幫助,我將不勝感激。
斯威夫特
@IBAction func modPres(_ sender: AnyObject) {
let postDataURL = "https://www.juankarfollador.com/login_app.php"
let url: NSURL = NSURL(string: postDataURL)!
let request: NSMutableURLRequest = NSMutableURLRequest(url:url as URL)
request.httpMethod = "POST"
request.httpBody = user.data(using: String.Encoding.utf8)
request.httpBody = l_origen.data(using: String.Encoding.utf8)
request.httpBody = l_destino.data(using: String.Encoding.utf8)
request.httpBody = num_pal.data(using: String.Encoding.utf8)
request.httpBody = String(precio).data(using: String.Encoding.utf8)
request.httpBody = texto.data(using: String.Encoding.utf8)
NSURLConnection.sendAsynchronousRequest(request as URLRequest, queue: OperationQueue.main)
{
(response, data, error) in
print(response!)
if let httpResponse = response as? HTTPURLResponse {
let statusCode = httpResponse.statusCode
if statusCode==200 {
print("Connection Successful")
} else {
print("Connection Failed (!200)")
}
}
}
}
腓
$precio = $_POST['precio'];
$texto = $_POST['texto'];
$user = $_POST['user'];
$l_origen = $_POST['l_origen'];
$l_destino = $_POST['l_destino'];
$num_pal = $_POST['num_pal'];
$modificar = $_POST['modificar'];
define('HOST','***');
define('USER','***');
define('PASS','***');
define('DB','***');
$con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect');
mysqli_set_charset($con, 'utf8');
//Cliente
$sql = "UPDATE users SET precio='$precio', text_cli='$texto', l_origen='$l_origen', l_destino='$l_destino', num_pal='$num_pal' WHERE username='$user' AND text_cli=''";
mysqli_query($con,$sql);
控制檯打印 「連接成功」,這就是爲什麼我認爲連接做得很好(我不知道,雖然,因爲我對斯威夫特很新穎)
感謝您的時間@meda。該PHP頁面只是爲了測試它。當它工作時,我會製作更安全的腳本。我會嘗試你的代碼並告訴你結果。再次感謝你。 – alberzyzz
@alberzyzz記住它沒有測試,我的觀點是構建paramString,並且您可以刪除編碼,因爲我在我的答案的最後一行做了編碼 – meda
未測試但工作!非常感謝你。 – alberzyzz