(isset($_POST['submitted'])) doesn't seem to work - php

Question

I can't figure out what I'm doing wrong here. I've been watching different videos on this problem, and been searching for answers for a long time.
When I hit the submit button on the web page, the fields get cleared and nothing else happens. There is probably a simple mistake, but I can't find out what I'm doing wrong
<a1>
<h1>Registrer ny bruker</h1>
<hr>
<form method="post" action="" >
<input type="text" name="surName" placeholder="Etternavn" required>
<br><br>
<input type="text" name="givenName" placeholder="Fornavn" required>
<br><br>
<select name="gender">
<option value="male">male</option>
<option value="female">female</option>
</select>
<br><br>
<input type="number" name="age" placeholder="Alder" required>
<br><br>
<input type="text" name="mail" placeholder="e-post" required>
<br><br>
<input type="text" name="userName" placeholder="Brukernavn" required>
<br><br>
<input type="text" name="password" placeholder="Passord" required>
<br><br>
<input type="submit" name="submitted" value="Registrer">
</form>
</a1>
<?php
if(isset($_POST['submitted'])){
$mysqli = new mysqli('my database username and password');
$email = $mysqli->real_escape_string($_POST['mail']);
$surName = $mysqli->real_escape_string($_POST['surName']);
$givenName = $mysqli->real_escape_string($_POST['givenName']);
$username = $mysqli->real_escape_string($_POST['userName']);
$password = md5($_POST['password']);
$gender = $mysqli->real_escape_string($_POST['gender']);
$age = $mysqli->real_escape_string($_POST['age']);
$sql = "INSERT INTO blog_user (first_name, last_name, age, gender, user_password, user_name, mail)
VALUES ('$givenName','$surName','$age','$gender','$password','$username','$email')
";
$insert = $mysqli->query($sql);
if ($insert) {
echo "Succes!";
} else {
die("Error: {$mysqli->errno} : {$mysqli->error}");
}
$mysqli->close();
}
?>
Edit : I made a small typo when I first added this, so the first answers did not fix my problem. I did not give them a down vote, and I'm sorry I made a typo.

Answer 1

<?php
if(isset($_POST['submitted'])){
$mysqli = new mysqli('my database username and password');
$email = $mysqli->real_escape_string($_POST['mail']);
$surName = $mysqli->real_escape_string($_POST['surName']);
$givenName = $mysqli->real_escape_string($_POST['givenName']);
$username = $mysqli->real_escape_string($_POST['userName']);
$password = md5($_POST['password']);
$gender = $mysqli->real_escape_string($_POST['gender']);
$age = $mysqli->real_escape_string($_POST['age']);
$sql = "INSERT INTO blog_user (first_name, last_name, age, gender, user_password, user_name, mail)
VALUES ('$givenName','$surName','$age','$gender','$password','$username','$email')
";
$insert = $mysqli->query($sql);
if ($insert) {
echo "Succes!";
} else {
die("Error: {$mysqli->errno} : {$mysqli->error}");
}
$mysqli->close();
}
?>
you have missed if statement.

Answer 2

Your code is "clearing the fields" because your form is always presented first with default values. The default values are not given with the input fields, so it's therefore "cleared" everytime you execute your php-file.
<form method="post" action="" >
<input type="text" name="surName" placeholder="Etternavn" required>
<br><br>
<input type="text" name="givenName" placeholder="Fornavn" required>
<br><br>
<select name="gender">
<option value="male">male</option>
<option value="female">female</option>
</select>
<br><br>
<input type="number" name="age" placeholder="Alder" required>
<br><br>
<input type="text" name="mail" placeholder="e-post" required>
<br><br>
<input type="text" name="userName" placeholder="Brukernavn" required>
<br><br>
<input type="text" name="password" placeholder="Passord" required>
<br><br>
<input type="submit" name="submitted" value="Registrer">
</form>
For this reason you should use the form-request-part before you show the form
if(isset($_POST['submitted'])){
$mysqli = new mysqli('my database username and password');
$email = $mysqli->real_escape_string($_POST['mail']);
$surName = $mysqli->real_escape_string($_POST['surName']);
$givenName = $mysqli->real_escape_string($_POST['givenName']);
$username = $mysqli->real_escape_string($_POST['userName']);
$password = md5($_POST['password']);
$gender = $mysqli->real_escape_string($_POST['gender']);
$age = $mysqli->real_escape_string($_POST['age']);
$sql = "INSERT INTO blog_user (first_name, last_name, age, gender, user_password, user_name, mail)
VALUES ('$givenName','$surName','$age','$gender','$password','$username','$email')
";
$insert = $mysqli->query($sql);
if ($insert) {
echo "Succes!";
} else {
die("Error: {$mysqli->errno} : {$mysqli->error}");
}
$mysqli->close();
}
and show default values something like this...
<form method="post" action="" >
<input type="text" name="surName" placeholder="Etternavn"
required value="<?php echo $surName;?>">
<br><br>
<input type="text" name="givenName" placeholder="Fornavn"
required value="<?php echo $givenName;?>">
etc...
NOTE Database-query should be using prepared statements to avoid sql
injections.

Answer 3

Try this instead...
<?php
if(isset($_POST['submitted'])){