Ajax post json response issue - php

Question

I have an issue with ajax response. I am using custom query for fetch result from database. Json response always shows null value while query is running successfully. Here is my code:
if(isset($_POST)){
$arr = array();
//$query = mysql_query(" insert into login (user,pass) values ('".$_POST['firstname']."','".$_POST['lastname']."') ") or die('test');
$query = mysql_query(" select * from login where user = '".$_POST['firstname']."' && pass = '".$_POST['pass']."' ") or die('test');
$rows = mysql_num_rows($query);
while($fetch = mysql_fetch_array($query))
{
if($fetch)
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
else
{
$arr['failed']= "Login Failed try again....";
}
}
echo json_encode($arr);
}

Answer 1

#Amandhiman i did not get what is the use of if statement with in the while
if($fetch)
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
the mention code definitely works for you
if($rows>0)
{
while($fetch = mysql_fetch_array($query))
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
}else
{
$arr['failed']= "Login Failed try again....";
}

Answer 2

Try with the below code, (mysql is deprected), working for me with my test database table (Debug it, var_dump $result, $result->fetch_assoc(), $result->num_rows)
<?php
$servername = "localhsot";
$username = "yourser";
$password = "passyour";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
if(isset($_POST)){
$arr = array();
$query = "select * from login where user = '".$_POST['firstname']."' && pass = '".$_POST['pass']."' ";
$result = $conn->query($query);
$rows = $result->num_rows;
while($fetch = $result->fetch_assoc())
{
if($fetch)
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
else
{
$arr['failed']= "Login Failed try again....";
}
}
echo json_encode($arr);
}

Answer 3

try this
if(isset($_POST)){
$arr = array();
//$query = mysql_query(" insert into login (user,pass) values ('".$_POST['firstname']."','".$_POST['lastname']."') ") or die('test');
$query = mysql_query(" select * from login where user = '".$_POST['firstname']."' && pass = '".$_POST['pass']."' ") or die('test');
$rows = mysql_num_rows($query);
if($rows>0)
{
while($fetch = mysql_fetch_array($query))
{
$_SESSION['ID']= $fetch['id'];
$arr['id'] = $_SESSION['ID'];
}
}else
{
$arr['failed']= "Login Failed try again....";
}
echo json_encode($arr);
}

Answer 4

First of all start session before playing with it on the top of page
session_start();
check your database connectivity.
Use print_r($arr) for testing your array();

Answer 5

first off all you are using session variable . to use session variable you need to initialize it by session_start()
you are using key in the array in this way it will return the last inserted record in the array . try this code
<?php
$servername = "localhsot";
$username = "yourser";
$password = "passyour";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error)
die("Connection failed: " . $conn->connect_error);
if(isset($_POST)){
$arr = array();
$query = "select * from login where user = '".$_POST['firstname']."' && pass = '".$_POST['pass']."' ";
$result = $conn->query($query);
$rows = $result->num_rows;
while($fetch = $result->fetch_assoc())
{
if($fetch)
{
// if wants to use session then start session
session_start(); // else it will return null
$_SESSION['ID']= $fetch['id']; // i don't know why this ??
// $arr['id'] = $_SESSION['ID']; comment this
$arr[] = array('msg'=>'succsee','ID'=>$fetch['id']);
}
else
{
$arr[]= array('msg'=>'fail','ID'=>null);
}
}
echo json_encode($arr);
}