注意：未定义的索引错误[重复]

Question

This question already has an answer here:

• “Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP 28 answers

I'm getting this error:

Notice: Undefined index: prospectname in C:\wamp\www\var\SAS\insert.php on line 12

Here is the code

<form action="#" method="POST" class='form-horizontal form-bordered'>
    <div class="control-group">
        <label for="prospectname" class="control-label">Prospect Name</label>
        <div class="controls">
            <input type="text" name="prospectname" id="prospectname" placeholder="prospectname" class="input-xlarge">
        </div>
    </div>
    <div class="control-group">
        <label for="status" class="control-label">Status</label>
    </div>
</form>

sas.js

$(".registeradd").click(function () {

    var prospectname = $("#prospectname").val();
    var status = $("#status").val();
    var dataString = 'prospectname =' + prospectname + '&status=' + status;

    if (prospectname == '') {
        alert("Please Enter Some Text");
    } else {
        $("#flash").show();
        $("#flash").fadeIn(400).html;

        $.ajax({
            type: "POST",
            url: "insert.php",
            data: dataString,
            cache: false,
            success: function (html) {
                $("#display").after(html);
                $("#flash").hide();
            }
        });
    }
    return false;
});

Insert.php

<?php

$dbHost = 'localhost'; // usually localhost
$dbUsername = 'root';
$dbPassword = '';
$dbDatabase = 'test';

$db = mysql_connect($dbHost, $dbUsername, $dbPassword) or die ("Unable to connect to Database Server.");

mysql_select_db ($dbDatabase, $db) or die ("Could not select database.");

$prospectname =$_POST['prospectname'];
$status=$_POST['status'];

$sql_insert="insert into salesactivity(prospectname,status) values '$prospectname','$status')";

print $sql_insert;
mysql_query($sql_insert);

?>

</div>

Answer 1

Change your datastring

var dataString = 'prospectname ='+ prospectname +'&status='+status;

to a json object

var dataString = { 'prospectname': prospectname, 'status': status};