Converting my deprecated mysql to mysqli - php

Question

I'm a beginner here I need to learn the basic in converting from mysql to mysqli. I want to start in configuration first. I have used this connect.php and it's working now I wanted to switch to undeprecated mysqli. Please advise me how to modify this script.
<?php
$host = "localhost";
$dbusername = "root";
$dbpassword = "nopassword";
$dbname = "student";
$link_id = mysql_connect($host, $dbusername, $dbpassword);
if(!$link_id){
die(mysql_error("Can`t Connect To database"));
}
else{
$db = mysql_select_db($dbname, $link_id);
}
if(!$db){
die(mysql_error("Can`t select database"));
}
return;
?>

Answer 1

According to my comment....
you could do it that way
$host = "localhost";
$dbusername = "root";
$dbpassword = "nopassword";
$dbname = "student";
$link = mysqli_connect($host,$dbusername,$dbpassword,$dbname) or die("Error " . mysqli_error($link));
$query = "SELECT knowhow FROM google WHERE myknowhow='leaks'";
$knowhow = mysqli_query($link, $query);
this is just the simplest way you can do it.
But you should use it like this (OOP)
$db = new mysqli($host,$dbusername,$dbpassword,$dbname);
$knowhow = $db->query("SELECT knowhow FROM google WHERE myknowhow='leaks'");
According to your comment:
You dont need the IF/ELSE statements anymore. All errors are catched with
die("Error " . mysqli_error($link));
In the oop, errors are catched in
... new mysqli($host,$dbusername,$dbpassword,$dbname);
EDIT
Since you where blocked asking new questions ill just update this answere:
$link = mysqli_connect($host,$dbusername,$dbpassword,$dbname) or die("Error " . mysqli_error($link));
$query = "SELECT * FROM student_information where student_id='{$_SESSION['user_id']}'";
$knowhow = mysqli_query($link, $query);
$data = mysqli_fetch_array($result);
$numRows = mysqli_num_rows($result);
$i = 0;
while($i < $numRows){
echo $data[$i++] . "<br />";
}
// I would not use a while in this case. I would prefere a foreach
// just to prevent from using a endless loop and 1 row less code :)
foreach($data as $d){
echo $d . "<br />";
}
Please use MySqli as a Class like $db = new Mysqli();