如果sql数据库存在

Here is my code, this is just for school assignment and I know it is vulnerable to sql injection. I need if statement for creating sql database, so if already exists just skip it, here is the code:

<?php
if(isset($_POST['submit'])){
    $ime = $_POST['ime'];
    $prezime = $_POST['prezime'];
    $email = $_POST['email'];
    $adresa = $_POST['adresa'];
    $komentar = $_POST['komentar'];

        $conn=mysqli_connect('localhost','root','');
        mysqli_query($conn,"create database osobni_podatci");
        mysqli_select_db($conn,"osobni_podatci");
        mysqli_query($conn,"create table podatci(id int primary key auto_increment,ime varchar(50),prezime varchar(50),email varchar(20),adresa varchar(50),komentar varchar(100))");

        $conn=mysqli_connect('localhost','root','','osobni_podatci');
        $sql="INSERT INTO podatci(ime, prezime,email,adresa,komentar) VALUES('$ime','$prezime','$email','$adresa','$komentar')";
        mysqli_query($conn,$sql);
        mysqli_close($conn);
                echo "Podatci su uspjesno uneseni u bazu podataka 'osobni_podatci'";

            }elseif(isset($_POST['brisi'])){ 

        $con = mysqli_connect("localhost","root","","osobni_podatci");
        $delete = "DELETE FROM podatci";
        mysqli_query($con,$delete);
        mysqli_close($con);
                echo "Podatci su uspjesno izbrisani iz baze podataka.";

} ?>

mysqli_query($conn,"create database if not exists osobni_podatci");

You can use the query like

mysqli_query($conn,"create database if not exists osobni_podatci");

https://dev.mysql.com/doc/refman/5.0/en/create-database.html

Wrong line $conn=mysqli_connect('localhost','root',''); You have to enter 4 parameters first is host, second is login third is password and fourth is the database name