echoing wrong else in php - php

Question

Whenever I input wrong username, the resulting page gives the second else output. I want the first else to be displayed on the screen if a user types in wrong username and the second else to be displayed when someone tries to go in the login page directly from the url without inputting any name. And yes session_start(); has been declared on top of both the pages.
<?php
if (isset($_POST["submit"]))
{
$username = $_POST["username"];
$conn = new mysqli("localhost", "root", "", "test");
$result = $conn->query("select name from students where name = '$username'");
if ($result->num_rows > 0)
{
$_SESSION["username"] = $username;
echo "You are logged in. This is the welcome page. Welcome user: " . $username;
}
else
{
echo "Invalid username. Try again.";
}
$conn->close();
}
else
{
echo "Come through proper ways.";
}
?>

Answer 1

Possible issues
In general, you omitted some error management that could lead to unexpected behavior, which breaks the logic of your conditions.
You must check $_POST['username'], consider possible to receive $_POST['submit'] without an username (the web is full of surprises). The best way to differentiate missing username and bad username is to check it directly with isset() and empty() for instance.
You must check that the database connection succeeded to avoid exceptions with conn->connect_errno.
You must check if $result evaluates to false which would mean that there is a query error.
You may escape $username before inserting it into the request, I don't know how mysqli manages SQL injections.
Possible solution
<?php
if ( isset($_POST['submit']) && isset($_POST['username']) && !empty($_POST['username']) ) {
$conn = new mysqli("localhost", "root", "", "test");
$username = mysqli_real_escape_string($conn, $_POST["username"]);
// check connection
if ( $conn->connect_errno ){
die("Data access denied ([".$conn->connect_errno."] ".$conn->connect_error.")");
}
$result = $conn->query("select name from students where name = '$username'");
// check for query errors
if (!$result) {
die("Data access denied (invalid query)");
}
// ...
} else {
echo "Come through proper ways.";
}

Answer 2

try with following codes
<?php
if(isset($_POST["submit"] && array_filter($_POST) && $_POST['username']){
$conn = new mysqli("localhost", "root", "", "test") or die('Database Connection Failure ' . (($conn->connect_errno) ? $conn->connect_error : ''));
$username = mysqli_real_escape_string($conn, $_POST['username']);
$result = $conn->query("select name from students where name = '{$username}'");
if($result->num_rows > 0){
// success
$_SESSION["username"] = $username;
echo "You are logged in. This is the welcome page. Welcome user: " . $username;
}else{
echo "Invalid username. Try again.";
}
$conn->close();
}else{
echo "Come through proper ways.";
}