Why ajax code to connect with php is not working?

Question

I wrote the ajax in the JavaScript function. That code is

function getValidate(checkID)
{
    alert(checkID);
    $.ajax({                  
        type: 'post',
        url: 'checkval.php',
        datatype: 'json',
        data: {checkID : checkID},
        success: function (response) {

          if (response === "OK"){
            alert("Validation Successed.");

        }else if(response === "NG"){
            alert("Check Already Exists.");
        }
    },
        error : function(err, req) {
        alert("Error Occurred");
    }
  });
 }

this code is outputs only "Error Occurred".

the connected php script is

<?php
        echo("welcome");
    $check          = $_POST['checkID'];
    $host       = 'localhost';
        $database   = 'database';
        $username   = 'root';
        $password   = 'root';

    $dbc = mysqli_connect($host,$username,$password,$database);

        $checkno = $check;
        $sql = "select claimno from check_details where checkno = $checkno";
        $result = mysqli_query($dbc,$sql);
        $rows = mysqli_num_rows($result);
        if($rows != 0)
        {
                        echo "NG";
        }
                else
                {
                        echo "OK";
                }   
?>

at a time of calling the JavaScript function php file not executed......

please give me the idea to success it...........

What errors do you get? – evolutionxbox Aug 14 '15 at 08:41 — evolutionxbox, Aug 14 '15 at 08:41
i got "Error Occurred" this as a output... – Sathish Kumar D Aug 14 '15 at 08:43 — Sathish Kumar D, Aug 14 '15 at 08:43
Could you include the error message into your question? – evolutionxbox Aug 14 '15 at 08:44 — evolutionxbox, Aug 14 '15 at 08:44
use json_encode() in your php and return it – aldrin27 Aug 14 '15 at 08:44 — aldrin27, Aug 14 '15 at 08:44
where...... into a start of the php file...??? – Sathish Kumar D Aug 14 '15 at 08:46 — Sathish Kumar D, Aug 14 '15 at 08:46
Check your " url: 'checkval.php'" file name or file path – Tushar Aug 14 '15 at 08:59 — Tushar, Aug 14 '15 at 08:59

score 2 · Answer 1 · answered Aug 14 '15 at 08:50

2

Try this :

if($rows != 0)
{
    $return =  "NG";
}
else
{
    $return = "OK";
}   


echo json_encode($return);

answered Aug 14 '15 at 08:50

Happy Coding

2,517
1
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1

Please check path of checkval.php – Happy Coding Aug 14 '15 at 09:04
both are in a same directory – Sathish Kumar D Aug 14 '15 at 09:07
1

error: function(ts) { alert(ts.responseText); } Try this – Happy Coding Aug 14 '15 at 09:14
is to handle ajax error. Please try this and let me know – Happy Coding Aug 14 '15 at 09:21
it shows this error...."Sorry. you are not allowed access to this resource"... may be i found the exact error..... that's a user privilege error......... – Sathish Kumar D Aug 14 '15 at 09:26

score 1 · Answer 2 · edited May 23 '17 at 10:27

1

You are getting error occurred, because of below code. Try logging something meaningful to triag this.

error : function(err, req) {
        alert("Error Occurred");
    }

Please try below code to get a clue of the error

error: function(xhr, status, error) {
  var err = eval("(" + xhr.responseText + ")");
  alert(err.Message);
}

Reference: Take a look at this query

edited May 23 '17 at 10:27

Community

1
1

answered Aug 14 '15 at 08:52

Gyanendra Dwivedi

5,511
2
27
53

Jack jdeoel · Accepted Answer · 2017-07-13T01:42:31.920

Also you set datatype to json so response data must be json type

$.ajax({ dataType:"json"});

In php, store result in one variable and return json_encode

<?php
    echo("welcome");
$check          = $_POST['checkID'];
$host       = 'localhost';
    $database   = 'database';
    $username   = 'root';
    $password   = 'root';

$dbc = mysqli_connect($host,$username,$password,$database);

    $checkno = $check;
    $sql = "select claimno from check_details where checkno = '$checkno'";  //use single quote 
    $result = mysqli_query($dbc,$sql);
    $rows = mysqli_num_rows($result);
    if($rows != 0)
    {
                    $res = "NG";
    }
    else
    {
                    $res = "OK";
    }   
echo json_encode($res);  
?>

score 0 · Answer 4 · answered Aug 14 '15 at 08:45

0

It seems that the php script is not working well.

Debug it with try and catch and see what output is comming.

answered Aug 14 '15 at 08:45

Rachel Geller

302
1
3
12

php script is working without any error or warnings. the complete problem is ajax...... it's not connected with php script...... – Sathish Kumar D Aug 14 '15 at 08:51
Can you access the php script through a web browser? – Rachel Geller Aug 14 '15 at 08:53

Why ajax code to connect with php is not working?

4 Answers4