注意：未定义的索引替代修复

Question

<?php

    $db = mysqli_connect("localhost","root","","hell");
    mysql_connect("localhost", "root", "");
    mysql_select_db("accounts");
    $codenm = mysql_escape_string(($_POST['codenm']));

    $check = mysql_query ("SELECT codenm FROM users WHERE codenm='$codenm'");
    $check_num_rows = mysql_num_rows($check);

    if ($codenm == NULL)
        echo "Choose a Codename";
    else if (strlen($codenm)<4)
        echo "Too Short"; 
    else
    {
        if ($check_num_rows == 0)
        {
            echo "Available!";
        }else{
            if ($check_num_rows == 1)
            echo "Not Available";

    }
    }
?>

我已尝试isset，但当我在页面时echo只停留在

  

“太短”

当我使用isset

时，未显示不可用的回音和可用回音

但如果我没用过一个

我显示错误

  

注意：未定义的索引：第6行的codenm

当我输入codenm字段时，错误消失但回声不起作用