selecting 2 different option values and sending data to php using ajax - php

Question

I am taking two different values from select options and sending it using AJAX, but PHP is not responding to the request.
This is my jQuery code:
$(document).ready(function() {
$("#updateStatus").change(function() {
var opt = $("#updateStatus").val();
$("#updateStatus1").change(function() {
var sta = $("#updateStatus1").val();
$.ajax({
url: 'updatecode.php',
type: 'POST',
data: "option=" + opt + "&status=" + sta,
dataType: 'json',
success: function(data) {
alert(data + "hello");
}
});
});
});
});
This is my PHP code:
$id = $_POST['opt'];
$status = $_POST['sta'];
$query = "UPDATE projectstable SET projectStatus='".$status."'WHERE id='".$id."'";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
if (!$row) {
echo json_encode("fail");
} else {
echo json_encode("sucess");
}

Answer 1

Replace your existing ajax function like this
$(document).ready(function() {
var opt = '';
var sta = '';
$("#updateStatus").change(function() {
opt = $("#updateStatus").val();
});
$("#updateStatus1").change(function() {
sta = $("#updateStatus1").val();
callAjax(opt,sta);
});
});
function callAjax(opt,sta) {
$.ajax({
url: 'updatecode.php',
type: 'POST',
data: "opt=" + opt + "&sta=" + sta,
dataType: 'json',
success: function(data) {
alert(data + "hello");
}
});
}

Answer 2

You are passing the json body as a query string, and you are declaring to use dataType: "json"
$.ajax({
url: 'updatecode.php',
type: 'POST',
data: "option=" + opt + "&status=" + sta,
dataType: 'json',
success: function(data) {
alert(data + "hello");
}
});
instead that you should do it like this:
$.ajax({
url: 'updatecode.php',
type: 'POST',
data: {"option": opt, "status": sta},
dataType: 'json',
success: function(data) {
alert(data + "hello");
}
});

Answer 3

You are sending request under data using "option=" and "&status=" but you are reading these values in PHP through $_POST['opt'] and $_POST['stat'] which should actually be $_POST['option'] and $_POST['status'] respectively.
Also you need to change the data request to JSON format like data: {option:opt,status:sta} for your dataType is JSON

Answer 4

Your code should be like this
$(document).ready(function () {
$("#updateStatus,#updateStatus1").change(function () {
var opt = $("#updateStatus").val();
var sta = $("#updateStatus1").val();
$.ajax({
url: 'updatecode.php',
type: 'POST',
data: {option:opt,status:sta},
dataType: 'json',
success: function (data) {
alert(data + "hello");
}
});
});
})
and php code should like this
$id = $_POST['option'];
$status = $_POST['status'];
$query = "UPDATE projectstable SET projectStatus='".$status."'WHERE id='".$id."'";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
if (!$row) {
echo json_encode("fail");
} else {
echo json_encode("sucess");
}

Answer 5

Thanks every one looks like i just had to change some code in ajax this is my new code which is working fine
This is my PHP code I got rid of if(!$row)
mysql_select_db("dts_db",$con);
$id=$_POST['opt'];
$status=$_POST['sta'];
$query="UPDATE projectstable SET projectStatus='".$status."'WHERE id='".$id."'";
$result=mysql_query($query);if(!$result){ die("My sql query result ".mysql_error());}
else {
echo json_encode("success");
}
and this is my ajax
$.ajax({
url: 'updatecode.php',
type: 'POST',
data: {opt:opt,sta:sta},
dataType: 'json',
success: function(data) {
alert(data + "hello");
}
});