Hello together,
I'm still relatively new with PHP and I'm encountering an error with the error code (Warning: Undefined array key "title" in I:\xampp\htdocs\test\Datenspeichern.php on line 13 to 17.).
Normally I wanted that when I enter the data in the fields from the CD_Website page, that it is inserted in my created database and displayed in the table on the page.
Does anyone know where my error is?
Thanks a lot
main Code=
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Unbenanntes Dokument</title>
<style type="text/css">
table {
border-collaps: collapse;
width: 100%;
font-family: serif;
font-size: 35px;
text-align: center;
}
td {
font-size: 25px;
text-align: center;
font-family: serif;
}
</style>
</head>
<body>
<table>
<tr>
<th>CD-Titel</th>
<th>Artist</th>
<th>Songtitel</th>
<th>Musiklänge</th>
<th>Lied Nummer</th>
</tr>
<?php
$host = "localhost";
$user = "root";
$password = "";
$db_name = "cdaufgabe";
$con = mysqli_connect($host, $user, $password, $db_name);
if(mysqli_connect_error()) {
die("Verbindungsabbruch mit der Datenbank: ". mysqli_connect_error());
};
$check = "SELECT * FROM `cd-titel`";
$result = mysqli_query($con, $check);
if ($result > null) {
while ($row = $result->fetch_assoc()){
echo "<tr><td>" . $row['CD-Titel'] . "</td> <td>" . $row['Artist'] . "</td> <td>" . $row['Songtitel'] . "</td> <td>" . $row['Musiklänge'] . "</td> <td>" . $row['Lied-Nummer'] . "</td></tr>";
};
};
/*if (isset($_POST['submitted'])) {
$titel = $_POST['titel'];
$artist = $_POST['artist'];
$songtitel = $_POST['songtitel'];
$musicle = $_POST['musicle'];
$liednr = $_POST['liednr'];
$data_add = "INSERT INTO cd-titel (CD-Titel, Artist, Songtitel, Musiklänge, Lied-Nummer) VALUES ('$titel', '$artist', '$songtitel', '$musicle', '$liednr')";
if (!mysqli_query($db_name, $data_add)) {
die('Fehler beim einfügen von Daten');
}
}
*/
// $data_add = "INSERT INTO `cd-titel` (`CD-Titel`, `Artist`, `Songtitel`, `Musiklänge`, `Lied-Nummer`) VALUES ('$titel', '$artist', '$songtitel', '$musicle', '$liednr')";
// mysqli_query($con, $data_add);
/*mysqli_query($con, $data_add);*/
?>
<form methode="post" action="Datenspeichern.php">
<input type="text" name="titel" placeholder="CD-Titel"/>
<input type="text" name="artist" placeholder="Artist"/>
<input type="text" name="songtitel" placeholder="Songtitel"/>
<input type="text" name="musicle" placeholder="Musiklänge"/>
<input type="text" name="liednr" placeholder="Lied-Nummer"/>
<input type="submit" name="submitted" value="speichern"/>
</form>
</br></br></br>
</table>
</body>
</html>
second code(Datenspeichern.php)=
<?php
$con = mysqli_connect('localhost', 'root', '');
if (!$con) {
echo'Nicht verfügbar';
}
if (!mysqli_select_db($con, 'cdaufgabe')){
echo 'Datenbank nicht ausgewählt';
};
$titel = $_POST['titel'];
$artist = $_POST['artist'];
$songtitel = $_POST['songtitel'];
$musicle = $_POST['musicle'];
$liednr = $_POST['liednr'];
$data_add = "INSERT INTO cd-titel (CD-Titel, Artist, Songtitel, Musiklänge, Lied-Nummer) VALUES ('$titel', '$artist', '$songtitel', '$musicle', '$liednr')";
if (!mysqli_query($con, $data_add)){
echo 'Fehler';
}
else {
echo'Eingefügt';
};
header("url=CD_Webseite.php");
?>
Does anyone knows the mistake i keep doing?