How to return an error status on an ajax call in PHP?

Question

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?

You can make use of console to see any errors in AJAX or Javascript — Suyog, Sep 28 '15 at 05:49
Side note: this is not a good idea -> `$query=mysqli_query($con,"INSERT INTO job(job_name,job_desc,status) values('$jname','$jd','$jstat') ");` SQL-injection prone. — Rasclatt, Sep 28 '15 at 05:50
Whatever you echoed in addj.php , will catch in ajax response, so echoed possible errors in addj.php file. — Shahzad Barkati, Sep 28 '15 at 05:52
I think you can use the `mysqli_real_escape_string()` but the best is to look up bind and parametrized statements — Rasclatt, Sep 28 '15 at 05:58
@prettyme, using `succss:function(response){ alert(response); }`. And for better query use `PDO` instead of `mysqli_*` — Shahzad Barkati, Sep 28 '15 at 05:58
@SHAZ I think PDO being better is a matter of opinion. I personally like PDO better and use it exclusively, but I don't know that it's better. — Rasclatt, Sep 28 '15 at 05:59

score 2 · Answer 1 · answered Sep 28 '15 at 05:56

Use this

            $.ajax({
                type:"get",
                url:"addj.php?content=" + a3 + "," + a4 + "," + a5,
                success: function(data) { alert("succsess") },
                error: function (xhr, ajaxOptions, thrownError) {
                        alert(xhr.status + " : " + xhr.responseText);

                 }
            });

score 2 · Accepted Answer · answered Sep 28 '15 at 05:59

Use success or error callback function and output data to console.

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,
            success: function(dataReturn, textStatus, jqXHR) {
                console.log(dataReturn);
            },
            error: function(jqXHR,textStatus,errorThrown){
                console.log(textStatus,errorThrown);
            }
        });
     lod();
    }else{alert('Fill all fields')}
}

score 0 · Answer 3 · edited May 23 '17 at 12:30

0

You can use the console of the browser. Try following link that has step by step.

Why the Ajax script is not running on IIS 7.5 Win 2008 R2 server?

edited May 23 '17 at 12:30

Community

1
1

answered Sep 28 '15 at 05:53

Techie

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score 0 · Answer 4 · answered Sep 28 '15 at 06:00

0

You can use this code to catch any ajax errors.

$.ajax({
  type:"get",
  url:"addj.php?content=" + a3 + "," + a4 + "," + a5
}).fail(function(error){
  alert(error);
});

Hope it helps.

answered Sep 28 '15 at 06:00

Pablo Jomer

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How to return an error status on an ajax call in PHP?

4 Answers4