When call php from jquery via ajax ,have any response .I changed the dataTypeand put jsoninstead html.I´m thinking the issue is that,for the ajax call never trigger the php code,it seems $_POST['retriveForm'] never carries a value.
PHP:
if(isset($_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);
}
and jQuery :
$(document.body).on('click', '.edit', function() {
var id = $(this).data('id');
$.ajax({
type: "POST",
url: "is_the_same_page.php",
data: {
retriveForm: id
},
dataType: "json",
success: function(response) {
$('#myForm').find('input').eq(1).val(response.item1);
}
});
});
Code is all in the same page if that may be important.
Since the AJAX code is in the same script, make sure you don't output any of the normal HTML in this case. Your PHP code should be at the very beginning of the script, before any HTML is output. And you should exit the script after echoing the JSON. Otherwise your JSON will be mixed together with HTML, and jQuery won't be able to parse the response.
Also, you're not correctly adding to $data_json in your loop. You're overwriting the variable each time instead of pushing onto it.
<?php
// code here to set up database connection
if(isset($_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);
exit();
}
?>
<html>
<head>
...
Then in the success function, you'll need to index the elements of response, since it's an array, not a single object.