mysql doesn't update information from html/php - php

Question

The problem I have: I made a "contact" form, which should send data to my database. All works good, no errors after I access the page from localhost, shows the result I wanted to see, but the database (localhost/phpmyadmin/..) doesn't update with any info.
This is my PHP:
if(isset($_POST['insert']))
{
$hostname = 'localhost';
$username = 'root';
$password = '';
$databaseName = 'nig';
$Nume = $_POST['nume'];
$Email = $_POST['email'];
$Telefon = $_POST['telefon'];
$Subiect = $_POST['subiect'];
$Mesaj = $_POST['mesaj'];
$connect = mysqli_connect($hostname, $username, $password, $databaseName);
$query = "INSERT INTO `amar` (`nume`, `email`, `telefon`, `subiect`, `mesaj`) VALUES ('$Nume','$Email','$Telefon','$Subiect','$Mesaj')";
$result = mysqli_query($connect,$query);
if($result)
{
echo 'Mesaj trimis.';
}else{
echo 'Mesaj netrimis';
}
mysqli_free_result($result);
mysqli_close($connect);
This is my HTML:
<form action="insert2.php" action="post">
<form role="form">
<div class="form-group">
<label for="nume">Nume complet</label>
<input type="text" class="form-control" name="nume">
</div>
<div class="form-group">
<label for="email">Adresa e-mail</label>
<input type="text" class="form-control" name="email">
</div>
<div class="form-group">
<label for="telefon">Telefon</label>
<input type="text" class="form-control" name="telefon">
</div>
<div class="form-group">
<label for="subiect">Subiect</label>
<input type="text" class="form-control" name="subiect">
</div>
<div class="form-group">
<label for="mesaj">Mesaj</label>
<textarea class="form-control" name="mesaj" rows="8"></textarea>
</div>
<input type="submit" name="insert" class="btn btn-theme" value="insert"></button>
</form>
And the result should be data in my MySQL. but the database isn't getting any data, what am I doing wrong?

Answer 1

You have two form tags, one inside the other.
Change this....
<form action="insert2.php" action="post">
<form role="form">
To this...
<form action="insert2.php" action="post">
Also, please note, your PHP code is open to SQL injection. Never trust user input. Always validate and/or escape user entered data prior to using it in a query. Or better yet, used prepared statements or stored procedures.
Update 1:
OK, sorry to hear this is not working. Please add the following mysqli error debugging so we can see what is happening.
Change this...
$result = mysqli_query($connect,$query);
To this...
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$result = mysqli_query($connect,$query)
if (!$result)) {
printf("Errormessage: %s\n", mysqli_error($connect));
}
Then run the code and see if you get any errors that may help us.
Update 2:
OK, try another piece of debugging.
After this...
$query = "INSERT INTO `amar` (`nume`, `email`, `telefon`, `subiect`, `mesaj`) VALUES ('$Nume','$Email','$Telefon','$Subiect','$Mesaj')";
Add this...
die($query);
Save and run the code.
You will be given a SQL query. Copy and paste that SQL query and run it in your database management tool (phpMyAdmin for example). See if the record is added when you manually run the query or if you are given any errors.
Let us know the outcome.