CODEWITHSUNDEEP
phphtmlmysqldatabase
Stöger
Stöger

Reputation: 155

Storing form input in phpMyAdmin database with PHP

I'm trying to learn how to use databases in PHP.

My main html file creates somewhat like a login form. The data which was used as input there is redirected to the PHP file, where it should be printed and also stored in the database.

<?php error_reporting(E_ALL);

        function db_con($data_array){
            $db = mysqli_connect("localhost", "root", "", "form") or die("Fehler beim Herstellen der Verbindung zur Datenbank!");
            mysqli_set_charset($db, 'utf8');

            mysqli_query($db, "
                INSERT INTO 'data'
                (
                    'E-Mail', 'Name', 'Bewertung', 'Kommentar', 'Thema'
                )
                VALUES
                (
                    '".$data_array['E-Mail']."',
                    '".$data_array['Name']."',
                    '".$data_array['Buchbewertung']."',
                    '".$data_array['Kommentar']."',
                    '".$data_array['Lieblingsthema']."'
                )
            ") or die("Fehler beim Schreiben der Datensätze!");
        }

            if(isset($_POST['name'], $_POST['email'], $_POST['rating'], $_POST['comment'], $_POST['interest'])){
            $data_array = array(
                    "Name" => $_POST['name'],
                    "E-Mail" => $_POST['email'],
                    "Buchbewertung" => $_POST['rating'],
                    "Kommentar" => $_POST['comment'],
                    "Lieblingsthema" => $_POST['interest']
                );

            echo "<th colspan='2'>Folgende Daten wurden übermittelt: </th>";
            echo "<br><br>";

                foreach($data_array as $key => $val){
                    echo "<tr><td><b>". $key. ": ". "</b></td><td>". $val. "</td></tr>";
                }

                db_con($data_array);
            }
        ?>

I don't get why it doesn't work. There is no error, but the script stops after mysqli_query (last output: Fehler beim Schreiben der Datensätze!).

Can someone help me why all those data sets won't be stored in the database and how to fix it?

Upvotes: 1

Views: 1611

Answers (3)

Jhon Sebastian Lopez
Jhon Sebastian Lopez

Reputation: 1

I think that you don't have correct credentials from 'root', you should try to make the same but with other user of database.

Upvotes: 0

Kevin Morris
Kevin Morris

Reputation: 115

Guten Abend Stoger

Biggest problem with arrays is no output and debug Use this:

<pre>
  <?php print_r($_POST); print_r($_GET); ?>
</pre>

This will show you POST and GET values/ Also can use print_r($data_array). Use the pre tags so it formats. First step is to make sure something is in POST. I don;t see a form or where you get your user inputs from. Normally a form asks this and then you assign the POST values to variable

i.e. $name = $_POST['name']);

So add those lines in a php page and then see if array empty or not. Likely it is. If not then you need to see if the fail is mysql insert and you can set some debug there to fail if no insert. I am new but still your layout and detail can't be answered without more. Gruss!

Upvotes: 1

Zeshan Sajid
Zeshan Sajid

Reputation: 559

if(isset($_POST['name'], $_POST['email'], $_POST['rating']$_POST['comment'], $_POST['interest'])){

                    $name = $_POST['name']);
                    $email = $_POST(['email']);
                    $rating = $_POST['rating']);
                    $comment = $_POST(['comment']);
                    $interest = $_POST(['interest']);


$sql = "INSERT INTO data ". "(E-Mail, Name, Bewertung, Kommentar, Thema) " . "VALUES('$email', '$name', '$rating', '$comment', '$e', '$interest')";
mysql_select_db('DATABASE NAME');
            $retval = mysql_query($sql, $conn); //$conn for your connection to database

            if(! $retval) {die('Could not enter data: ' . mysql_error());}}
}

Upvotes: 0

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