使用phpmyadmin和SQL登录功能

Question

So i've made this login function using PHP and im having trouble asking a question correctly to the phpmyadmin database, basically the question i ask now is "SELECT * FROM User" which only lets the first registered user in the database login.

The question i really want to ask is something like this: fetch the column where the username thats being inputed matches with the column.

This is my take on it:

 "SELECT * FROM User WHERE username = '$_post['username_input']'"

but i just get error message.

How should i ask this question? thanks for the help in advance!

<form method='post'>
    <table>
        <tr><td></td><td><input type='text' name='username_input' /></td></tr>
        <tr><td></td><td><input type='text' name='password_input' /></td></tr>
        <tr><td></td><td><input type='submit' name='submit' value='Logga in' /></td></tr>
    </table>
<form>

<?php
    $link = mysqli_connect ("localhost", "xxxxxxxx", "xxxxxxx", "xxxxxxx");
    $sql_result = $link->query ("SELECT * FROM User");
    $rad = mysqli_fetch_array ($sql_result); 

    $columnuser = $rad['username'];
    $columnpassword = $rad['password'];

    if (isset($_POST['submit'])) {
        if ($_POST['username_input'] == $columnuser && md5 ($_POST['password_input']) == $columnpassword) {
            $_SESSION['log_in'] = true;
            header("Location: administration.php");
        }
        else {
            echo "Fel användarnamn eller lösenord, vänligen försök igen.";
        }
    }
    mysqli_close ($link);
?>

Answer 1

What I am thinking you are making mistake in query is concatenation of variable and $_POST is array which is capitalized

"SELECT * FROM User WHERE username = '".$_POST['username_input']."' "

I hope this solves your problem.

Edit: The error you might be getting because code runs always, you should have condition in starting as

<?php
    if (isset($_POST['submit'])) {    
        $link = mysqli_connect ("localhost", "xxxxxxxx", "xxxxxxx", "xxxxxxx");
        $sql_result = $link->query ("SELECT * FROM User WHERE username = '".$_POST['username_input']."' ");
        $rad = mysqli_fetch_array ($sql_result); 

        $columnuser = $rad['username'];
        $columnpassword = $rad['password'];


        if ($_POST['username_input'] == $columnuser && md5 ($_POST['password_input']) == $columnpassword) {
             $_SESSION['log_in'] = true;
             header("Location: administration.php");
        }
        else {
            echo "Fel användarnamn eller lösenord, vänligen försök igen.";
        }
        mysqli_close ($link);
    }
?>