Retrieve data via JSON and put in a form - php

Question

I´ve tried retrive some info from mysql for put it in a form,and update the info.
I use ajax call in the same page,I call ajax passing the id and php return all info.
if(isset($_POST["retriveForm"])) {
$data_json =array();
$id = $_POST['retriveForm'];
$sql = "SELECT * FROM mytable WHERE Id = $id";
while ($row = mysqli_fetch_array($db->consulta($sql)) {
$data_json = array('item1' => $row['item1'],'item2' => $row['item2']) ;
}
$data_json['item_array'] = call_a_function_return_array();//this works
echo json_encode($data_json);
}
ajax call shows like this
$(document.body).on('click', '.edit' ,function(){
var id = $(this).data('id');
$.ajax({
type: "POST",
url: "is_the_same_page.php",
data: {retriveForm : id},
success: function(response) {
$('#myForm').find('input').eq(1).val(response.item1);
}
});
});
When put .val(response.item1) get anything,but if I put response the input take all values
$('#myForm').find('input').eq(1).val(response);
shows
{"item1":"item1","item2":"item2","item_array":["item_in_array1","item_in_array2"]}
I also tried
var obj = jQuery.parseJSON(response);

Answer 1

Try setting $.ajax() dataType options to "json"
"json": Evaluates the response as JSON and returns a JavaScript
object.
$(document.body).on("click", ".edit", function() {
var id = $(this).data("id");
$.ajax({
type: "POST",
url: "is_the_same_page.php",
dataType: "json", // set `dataType` to `"json"`
data: {
retriveForm: id
},
success: function(response) {
$("#myForm").find("input").eq(1).val(response.item1);
}
});
});