MySQL注意：未定义的索引：如何解决[重复]

I am creating two level dropdown select box with PHP, MYSQL and JQuery. Everything seems fine but when my parent category doesn't have child or if I doesn't select child it is giving me error and I can't understand what this error means and how to resolve. I need you guyz help.

Error: Notice: Undefined index: f_child_cat in C:\xampp\htdocs\pathtofile\index.php on line 58 You selected 12 & You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

Here is my front end code

<?php if(isset($_POST['submit'])){
    $drop = $_POST['first_child_cat'];
    $f_child_cat = $_POST['f_child_cat'];
    echo "You selected ";
    echo $drop." & ".$f_child_cat;

    $query =    mysql_query("SELECT * FROM qa_categories WHERE parentid = $f_child_cat")
                or die(mysql_error());

                echo '<ul>';
                while($cats = mysql_fetch_array( $query ))
                    echo '<li>',$cats['title'],'</li>';

                echo '</ul>';
}
?>

</div>

It means that $_POST['f_child_cat'] is not defined (doesn't exist). You should be using binded parameters to prevent against SQL injection and use mysqli_ or PDO functions.

Since this is part of your query, I would recommend that you wrap your query like so:

if( isset($_POST['f_child_cat']) && !empty($_POST['f_child_cat']) ) {
    ... call db
}
else {
    echo "You didn't select this"
}

$_POST['f_child_cat'] is not defined check if the value is present first before using it.

$f_child_cat = isset($_POST['f_child_cat'])?$_POST['f_child_cat']:false;

if ($f_child_cat){
    $f_child_cat = mysql_real_escape_string($f_child_cat); // always escape

    /// your query.
}

Note: mysql_* functions are depricated, use mysqli or pdo as an alternative.