如何在PHP中的ajax调用中返回错误状态？

I have this script code and it works perfectly,

CODE

<script>
    function ajob(){
        var a3=a.value
        var a4=b.value
        var a5=c.value
        if(a3!='' && a4!=''){
            $.ajax({
                type:"get",
                url:"addj.php?content=" + a3 + "," + a4 + "," + a5
            });
         lod();
        }else{alert('Fill all fields')}
    }
</script>

<form>
    <table>
        <tr><td>Job Name:</td>
            <td><input id="a" name="jname" type="text"></td>
        </tr>
        <tr><td>Job Description:</td>
            <td><input id="b" name="jd" style="margin: 2px 0 2px 0;" type="text"></td>
        </tr>
        <tr><td>Status:</td>
            <td>
                <select id="c" name="jstat" style="width:100%; height: 26px" >
                    <option>Active</option>
                    <option>Not Active</option>
                </select>
            </td>
        </tr>
    </table>
</form>

what i want to know is how can can i know if there is an error in my addj.php file? Here is my addj.php

addj.php

require 'con.php';
$pieces = explode(",", $_GET['content']);
$jname=$pieces[0];
$jd=$pieces[1];
$jstat=$pieces[2];
$query=mysqli_query($con,"INSERT INTO job(job_name,job_desc,status) values('$jname','$jd','$jstat') ");

How do I return a failed status from my PHP code and how can I handle that error in my java script code?