使用php [duplicate]从mysql数据库中的特定用户获取信息

When i enter a user id that is in the database, i want it to echo/print all the data from that one users file. So if i enter "user_id = 2", i want the users "name", "age" etc etc. I got to where that it gives me everything without the parameters. The code:

$link = mysqli_connect("localhost", "root", "");
mysqli_select_db($link, "magicsever");

if(mysqli_connect_error()){
    die ("Database connection error");
}

$query = "SELECT * FROM classified_videos";
$result = mysqli_query($link, $query);
while($row = mysqli_fetch_array($result)){      
    print_r ($row);
}

But i want from a specific user and this is what i think is the code

if(isset($_POST['submit'])){

    $link = mysqli_connect("localhost", "root", "");
    mysqli_select_db($link, "magicsever");

    if(mysqli_connect_error()){     
        die ("Database connection error");
    }   

    $query = "SELECT * FROM classified_videos WHERE user_id ='".mysqli_real_escape_string($link, $_POST['userid'])."' LIMIT 1";
    $result = mysqli_query($link, $query);
    while($row = mysqli_fetch_array($result)){      
        print_r ($row['vid_category_1']);
    }
}
?>

<form method="post">
    <input type="text" name="userid" placeholder="user id...">
    <input type="submit" value="submit">
</form>

</div>

This if(isset($_POST['submit'])) expects that an input with name submit is submitted, but youre currently sending only userid

add this name="submit" to your button:

<input type="submit" value="submit" name="submit">