Getting Ajax reponse as null in php - php

Question

I have a form in that I am trying to do inline editing and adding using AJAX call.
Firstly I am displaying data in HTML table. And then if enter data into text boxes and click on add button record adding displaying data in HTML table. After I click edit button data showing in the textboxes fine.
But I am getting the ajax response as null.
I couldn't figure it out.
This is my AJAX code PHP file:
$(function() {
$(".scrollingTable tbody a").click(function() {
//debugger;
var link = $(this).attr('href');
var arr = link.split('=');
var id = arr[1];
//alert(id);
$.ajax({
url: "insertgr.php",
type: "POST",
data: {
cntid: id
},
success: function(datas) {
var data = $.parseJSON(datas);
$("#num").val(data.id);
$("#namegr").val(data.vndr_cntname);
$("#designation").val(data.designation);
$("#mobilegr").val(data.vndr_cntmobile);
$("#maildgr").val(data.vndr_cntmail);
}
});
});
});
$(function() {
$('.txtcbt a').click(function() {
debugger;
var cntname, designation, mobile, email, vndrid, id, cid;
cid = $("#num").val();
cntname = $("#namegr").val();
designation = $("#designation").val();
mobile = $("#mobilegr").val();
email = $("#maildgr").val();
vndrid = "<?php echo $selectid; ?>";
//alert(cid);
if (cntname == "" || designation == "" || mobile == "" || email == "") {
alert("fields should not be empty");
} else {
$.ajax({
url: "insertgr.php",
type: "POST",
data: {
id: cid,
name: cntname,
dgnation: designation,
mobileno: mobile,
emailid: email,
vid: vndrid
},
success: function(html) {
var dat = $.parseJSON(html);
alert(html);
alert("it came to success");
$("#num").val("");
$('#namegr').val("");
$('#designation').val("");
$('#mobilegr').val("");
$('#maildgr').val("");
}
});
}
});
});
This is file AJAX is calling:
<?php
require('Assests/connection/connection.php');
error_reporting(0);
$vcntlist = "";
if (!empty($_POST['cntid'])) {
$id = $_POST['cntid'];
$result = mysqli_query($conn, "SELECT `id`, `vndr_cntname`, `designation`, `vndr_cntmobile`, `vndr_cntmail`,`vndr_id` FROM `vndr_cntdtls`where id=$id");
$rowcount = mysqli_num_rows($result);
if ($rowcount > 0) {
$row = mysqli_fetch_array($result);
$vcntid = $row['id'];
$cntname = $row['vndr_cntname'];
$cntdesignation = $row['designation'];
$cntmobile = $row['vndr_cntmobile'];
$cntmail = $row['vndr_cntmail'];
}
}
if (!empty($_POST['name']) && !empty($_POST['dgnation']) &&
!empty($_POST['mobileno']) && !empty($_POST['emailid']) &&
!empty($_POST['vid'])) {
$id = $_POST['id'];
$name = $_POST['name'];
$degination = $_POST['dgnation'];
$mobile = $_POST['mobileno'];
$email = $_POST['emailid'];
$vndrid = $_POST['vid'];
if (empty($_POST['id'])) {
$query = mysqli_query($conn, "INSERT INTO `vndr_cntdtls`(`vndr_cntname`, `designation`, `vndr_cntmobile`, `vndr_cntmail`, `vndr_id`) VALUES ('$name','$degination','$mobile','$email',$vndrid)");
} else {
$update1 = mysqli_query($conn, "UPDATE `vndr_cntdtls` SET `vndr_cntname`='$name',`designation`='$degination',`vndr_cntmobile`='$mobile',`vndr_cntmail`='$email' WHERE id=$id") or die(mysqli_error($conn));
}
$result = mysqli_query($conn, "SELECT DISTINCT `id`, `vndr_cntname`, `designation`, `vndr_cntmobile`, `vndr_cntmail`,vc.vndr_id FROM `vndr_cntdtls` vc INNER JOIN vendors v ON vc.vndr_id=$vndrid") or die(mysqli_error($conn));
$rowcount = mysqli_num_rows($result);
if ($rowcount > 0) {
while ($row = mysqli_fetch_array($result)) {
$vcntid = $row['id'];
$cntname = $row['vndr_cntname'];
$cntdesignation = $row['designation'];
$cntmobile = $row['vndr_cntmobile'];
$cntmail = $row['vndr_cntmail'];
}
}
}
echo json_encode($row);
?>

Answer 1

There could be a lot of ways it is not working.
First:
You need to put a application/json in your php so it can be compatible with the browser you are using.
header('Content-Type: application/json');
echo json_encode($row);
Second:
Why are you doing a while loop and assigning it into an unused variable?
You can simplify it by doing:
$row = mysqli_fetch_array($result);
header('Content-Type: application/json');
echo json_encode($row);
Unless it is multiple rows then:
$rowcount = mysqli_num_rows($result);
$rows = array();
if ($rowcount > 0) {
while ($row = mysqli_fetch_array($result)) {
$rows[] = $row;
}
}
header('Content-Type: application/json');
echo json_encode($row);
Third
null values usually appears when a variable you are trying to use is not initialized. By having the error_reporting turned off it does not display the error.
error_reporting(true);
Fourth
Check also the logs for database error, I assume it has something to do with MySQL query not having to reach the $row initialization.
Fifth
I believe you need to have it fetched as associative array for it to be useful
$row = mysqli_fetch_assoc($result);