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HTMLページのPHPスクリプトにテキスト入力を投稿しようとしています。 iOSにエラーはなく、PHPはSQLデータベースに正常に投稿できます。何らかの理由で、iOSはphpスクリプトに値を正常に渡すことができません。HTTPリクエストでPHPに投稿できませんが、コードにエラーはありません
@IBOutlet weak var header: UILabel!
override func viewDidLoad() {
super.viewDidLoad()
//helps the return button work
self.txt1.delegate = self
self.txt2.delegate = self
self.txt3.delegate = self
self.txt4.delegate = self
self.txt5.delegate = self
// Do any additional setup after loading the view, typically from a nib.
}
override func didReceiveMemoryWarning() {
super.didReceiveMemoryWarning()
// Dispose of any resources that can be recreated.
}
//Text Boxes
@IBOutlet weak var txt1: UITextField!
@IBOutlet weak var txt2: UITextField!
@IBOutlet weak var txt3: UITextField!
@IBOutlet weak var txt4: UITextField!
@IBOutlet weak var txt5: UITextField!
//function to make return button work..
func textFieldShouldReturn(_ textField: UITextField) -> Bool {
txt1.resignFirstResponder()
txt2.resignFirstResponder()
txt3.resignFirstResponder()
txt4.resignFirstResponder()
txt5.resignFirstResponder()
return true
}
//button action
@IBAction func Submit(_ sender: AnyObject) {
let requestURL = URL(string: "*****")
//You should use `URLRequest` in Swift 3, mutability is represented by `var`
var request = URLRequest(url:requestURL!)
request.httpMethod = "POST"
//UITextField.text can be nil, you should treat nil cases
//(Generally avoid using forced unwrapping `!` as far as you can.)
let song = txt1.text ?? ""
let artist = txt2.text ?? ""
let album = txt3.text ?? ""
let year = txt4.text ?? ""
let genre = txt5.text ?? ""
//`song`,... are all Strings, you have no need to add `as String`
let songPost = "song=" + song
let artistPost = "&artist=" + artist
let albumPost = "&album=" + album
let yearPost = "&year=" + year
let genrePost = "&genre=" + genre
//You need to make a single data containing all params
//(Creating a concatenated String and getting `data` later would be another way.)
var data = Data()
data.append(songPost.data(using: String.Encoding.utf8)!)
data.append(artistPost.data(using: String.Encoding.utf8)!)
data.append(albumPost.data(using: String.Encoding.utf8)!)
data.append(yearPost.data(using: String.Encoding.utf8)!)
data.append(genrePost.data(using: String.Encoding.utf8)!)
request.httpBody = data
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=\(error)")
print(response)
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 = \(response)")
}
//print response
let responseString = String(data: data, encoding: .utf8)
print("responseString = \(responseString)")
}
task.resume()
コンソールは、非常に大きなメッセージを出してくれるが、それはステータスコードが200であるべきと言うが、私はHTML上にある送信ボタンをヒットするコードの一部を412
足りませんさページ？ここで
PHPスクリプトです：
<html>
<head>
<title>Information Gathered</title>
</head>
<body>
<?php
echo "<p>Data Processed!</p>";
$song = $_POST['song'];
$artist = $_POST['artist'];
$album = $_POST['album'];
$year = $_POST['year'];
$genre = $_POST['genre'];
echo $song . "</br>";
echo $artist . "</br>";
echo $album . "</br>";
echo $year . "</br>";
echo $genre . "</br>";
DEFINE ('DB_USER', '****');
DEFINE ('DB_PASSWORD', '****');
DEFINE ('DB_HOST', '****');
DEFINE ('DB_NAME', '****');
$dbc = @mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
OR die('Could not connect to MySQL: ' .
mysqli_connect_error());
$sql = "insert into music (song, artist, album, year genre)
values('$song', '$artist', '$album', '$year', '$genre',)";
$dbc->query($sql)
echo "<p> Data Entered!!!</p>"
?>
</body>
正確にはどういう意味ですか？私は2つの間に間違った通信があるかどうかを確認するためにPHPを追加しました、SQLデータベースは、アプリケーションを拒否している、またはアプリケーションは、PHPスクリプトを拒否ですか？ – xteetsx