JSON encode no output [closed] - php

Question

Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 4 years ago.
Improve this question
I've tried to look up different ways to get this json to output correctly but im not sure if im accessing the right variable in php/or the success function value.pTitle as well as how to I get the access to the other value out such as artTitle im failing somewhere not sure where or why.UPDATE fixed the php file added and an array $data[].
this is my php code.
$sqlPAQuery = "SELECT pTitle, GROUP_CONCAT(artTitle) AS
artTitle
FROM p
JOIN art ON art.pId = p.pId
GROUP BY pTitle";
if ($result=mysqli_query($conn,$sqlPAQuery))
{
$data = [];
while ($row=mysqli_fetch_array($result,MYSQLI_ASSOC))
{
$data[] = $row;
}
echo json_encode($data);
}
This is the outcome from the php encode of row:
[{"pTitle":"ent","artTitle":"11,12"},{"pTitle":"pro","artTitle":"10"},{"pTitle":"sports","artTitle":"1,13"}]
This is the html code:
<h3>Output: </h3>
<div id="output"></div>
<script id="source" language="javascript" type="text/javascript">
$(document).ready(function() {
$.ajax({
type: "POST",
url: 'Data.php',
data: "",
dataType: 'json',
success: function(data)
{
$.each(data, function(index, value) {
var pageTitle = value.pTitle; //get name
$('#output').append("<b>pageTitle: </b>"+pageTitle+"<br/>");
}
});
});
Output should be:
pageTitle: ent
pageTitle: pro
pageTitle: sport
FIXED THE PHP FILE WORKS

Answer 1

If you want to proces the json like that this is what you need to do:
if ($result=mysqli_query($conn,$sqlPAQuery)) {
$data = [];
while ($row=mysqli_fetch_array($result,MYSQLI_ASSOC))
{
$data[] = $row;
}
echo json_encode($data);
}