Getting error even though records inserting correctly - php

Question

Why I'm getting the error in my jquery even though my record inserting in database? I add some changes in this code, I convert it into mysqli function. But the problem is the error message comes out? Why is that?
The error message is came from below on my script
Here's my code
Script
$(document).ready(function(){
$("#save").click(function(){
ajax("save");
});
$("#add_new").click(function(){
$(".entry-form").fadeIn("fast");
});
$("#close").click(function(){
$(".entry-form").fadeOut("fast");
});
$("#cancel").click(function(){
$(".entry-form").fadeOut("fast");
});
$(".del").live("click",function(){
if(confirm("Do you really want to delete this record ?")){
ajax("delete",$(this).attr("id"));
}
});
function ajax(action,id){
if(action =="save")
data = $("#userinfo").serialize()+"&action="+action;
else if(action == "delete"){
data = "action="+action+"&item_id="+id;
}
$.ajax({
type: "POST",
url: "ajax.php",
data : data,
dataType: "json",
success: function(response){
if(response.success == "1"){
if(action == "save"){
$(".entry-form").fadeOut("fast",function(){
$(".table-list").append("<tr><td>"+response.cat_name+"</td><td>"+response.cat_code+"</td><td>"+response.letter+"</td><td><a href='#' id='"+response.row_id+"' class='del'>Delete</a></td></tr>");
$(".table-list tr:last").effect("highlight", {
color: '#4BADF5'
}, 0000);
});
$(".entry-form input[type='text']").each(function(){
$(this).val("");
});
}else if(action == "delete"){
var row_id = response.item_id;
$("a[id='"+row_id+"']").closest("tr").effect("highlight", {
color: '#4BADF5'
}, 0000);
$("a[id='"+row_id+"']").closest("tr").fadeOut();
}
}else{
alert("unexpected error occured, Please check your database connection");
}
},
error: function(res){
alert("Unexpected error! Try again.");
}
});
}
});
AJAX
<?php
error_reporting(0);
$mysqli = new mysqli("localhost", "root", "", "2015");
if(isset($_POST) && count($_POST)){
$cat_name = $mysqli->real_escape_string($_POST["cat_name"]);
$cat_code = $mysqli->real_escape_string($_POST["cat_code"]);
$letter = $mysqli->real_escape_string($_POST["letter"]);
//$phone = mysql_real_escape_string($_POST['phone']);
$item_id = $_POST['item_id'];
$action = $_POST['action'];
if($action == "save"){
$result = $mysqli->query("insert into category values('','".$cat_name."','".$cat_code."','".$letter."')");
$lid = mysqli_insert_id();
if($lid){
echo json_encode(
array(
"success" => "1",
"row_id" => $lid,
"cat_name" => htmlentities($cat_name),
"cat_code" => htmlentities($cat_code),
"letter" => htmlentities($letter),
//"phone"=>$unique_id,
)
);
}else{
echo json_encode(array("success" => "0"));
}
}
else if($action == "delete"){
//echo "delete from info where id = '".$item_id."'";
$res = $mysqli->query("delete from category where id = '".$item_id."'");
if($res){
echo json_encode(array( "success" => "1","item_id" => $item_id));
}else{
echo json_encode(array("success" => "0"));
}
}
}else{
echo json_encode(array("success" => "0"));
}
?>

Answer 1

The Reponse returned in success block was zero because this statement here $lid = mysqli_insert_id(); returned nothing.
Replace $lid = mysqli_insert_id(); with $mysqli->insert_id , as this is the right syntax to get the id of the recently added/inserted row.
Happy Coding:)

Answer 2

I got it. mysqli_insert_id() needs the connection of the database.
It should be like this
mysqli_insert_id($mysqli);