Post data using AJAX then retrieving the value of JSON array - php

Question

I am trying to create a validation to check if my data already exist or not using Ajax. It's works, but now I want to add JSON inside my Ajax php file and retrieve the array inside of the JSON, unfortunately it doesn't work as expected.
Take a look at my JQuery below :
$("#button").click(function(e) {
e.preventDefault();
var tempcode = $('#someobject').val();
$.ajax({
method:"POST",
dataType: "text",
url:'example.php',
data:{
Code: tempcode
},
success:function(data){
var my = JSON.parse(data);
alert(my.Title);
}
});
});
And below is example.php :
$AJAXCode = $_POST['Code'];
$myObj = array();
$strSQL = mysql_query("select * from sometable where Code ='$AJAXCode'");
$check = mysql_num_rows($strSQL);
if ($AJAXCode == NULL) {
$myObj->Title= "Choose Something";
$myObj->Total = $check;
$myJSON = json_encode($myObj);
echo $myJSON;
} else {
if ($check != 0) {
$myObj->Title= "Already Exist !";
$myObj->Total = $check;
$myJSON = json_encode($myObj);
echo $myJSON;
} else {
$myObj->Title= "You are good !!!";;
$myObj->Total = $check;
$myJSON = json_encode($myObj);
echo $myJSON;
}
}
As you can see at my JQuery scripts, I am trying to call "Title" inside my JSON array, but it doesn't work. Am I missing something ?
This is the result in Console.log, no error show up just this:

Answer 1

Firstly, you initialize your object the wrong way. Use instead of $myObj = array(); one of the following:
$myObj = (object) array();
or
$myObj = new StdClass();
For more information and more ways to create your object, look up this link.
Next, change the dataType attribute to:
dataType: "json"
in order to tell your script that your PHP response is encoded as JSON. You can use now your response the regular way, with JSON.parse(), try:
console.log(JSON.parse(data));

Answer 2

Replace Below Code :
$("#button").click(function(e) {
e.preventDefault();
var tempcode = $('#someobject').val();
$.ajax({
method:"POST",
dataType: "json",
url:'example.php',
data:{
Code: tempkode
},
success:function(data){
console.log(data.Title); // Example
}
});
});

Answer 3

Change variable tempkode to tempcode in data