Input information to database, display it with PHP - php

Question

I have a form where I enter text, then it goes to the database and after I want it to be immediately displayed on the same page. I enter the info and submit it, page reloads and nothing appears in database or page. Any ideas maybe?
Saving part:
if(isset($_POST['ok'])){
$Vardas = $_POST['vardas'];
$Epastas = $_POST['epastas'];
$Kam = $_POST['kam'];
$Zinute = $_POST['zinute'];
$Date = date('Y-m-d H:i:s');
/*$IP = $_SERVER[REMOTE_ADDR];*/
}
//else {die ("Neuzpildyta forma");}
$sql = "INSERT INTO table1 (vardas, epastas, kam, data, zinute)
VALUES ('$Vardas', '$Epastas','$Kam', '$Date', '$Zinute')";
//if (mysqli_query($dbc, $sql)) echo "Įrašyta";
//else die ("Klaida įrašant:" .mysqli_error($dbc));
Form:
<form method='post' action="">
<div class="form-group col-lg-4">
<label for="vardas" class="control-label">Siuntėjo vardas:</label>
<input name='vardas' type='text' class="form-control input-sm">
</div>
<div class="form-group col-lg-4">
<label for="epastas" class="control-label">Siuntėjo e-paštas:</label>
<input name='epastas' id="epastas" type='email' class="form-control input-sm">
</div>
<div class="form-group col-lg-4">
<label for="kam" class="control-label">Kam skirta:</label>
<input name='kam' type='text' class="form-control input-sm">
</div>
<div class="form-group col-lg-12">
<label for="zinute" class="control-label">Žinutė:</label>
<textarea name='zinute' class="form-control input-sm"></textarea>
</div>
<div class="form-group col-lg-2">
<input type='submit' name='ok' value='siųsti' class="btnbtn-default">
</div>
</form>

Answer 1

It looks like you're missing your $mysqli connection. Your code is also very susceptible to mysql injection so here is my recommendation.
$mysqli = new mysqli("localhost", "username", "password", "database_name");
if(isset($_POST['ok'])){
$Vardas = $mysqli->real_escape_string($_POST['vardas']);
$Epastas = $mysqli->real_escape_string($_POST['epastas']);
$Kam = $mysqli->real_escape_string($_POST['kam']);
$Zinute = $mysqli->real_escape_string($_POST['zinute']);
$Date = date('Y-m-d H:i:s');
$my_insert_query = "INSERT INTO table1 (vardas, epastas, kam, data, zinute)
VALUES ('$Vardas', '$Epastas','$Kam', '$Date', '$Zinute')";
$insert = $mysqli->query($my_insert_query);
if($insert){
echo "Success!";
}else{
echo "error" . $mysqli->error;
}
}
The code above should work and prevent any sql injection.