Append function() is not working in jQuery Ajax of PHP response

Question

I am getting the array through PHP and I want to append in in table but I tried to make it from last 2 days and uses all the function which available on stackoverflow but it not working... Please developers , help me and teach me how i append out output data in table as below -

PHP Code :-

$result = $stmt->fetchAll();
 foreach($result as $data => $value) {

   //$QL QUERY HERE

   $data = array('name' => $value["name"], 'amount' => $amount, 'invoice' => $Invoice, 'response' => '200');
   echo json_encode($data);
 }
 exit();

My PHP response is :-

{"name":"AMAZON_FBA","amount":"1","invoice":"25","response":"200"}
{"name":"AMAZON_IN","amount":"12","invoice":"22","response":"200"} 
{"name":"FLIPKART","amount":"42","invoice":"08","response":"200"} 
{"name":"PAYTM","amount":"36","invoice":"03","response":"200"} 
{"name":"Replacement","amount":"0","invoice":"17","response":"200"}

Ajax:-

$.ajax({
.
.
  success: function (data) {
   var my_orders = $("table#salesReturnTB > tbody.segment_sales_return");

   $.each(data, function(i, order){
    my_orders.append("<tr>");
    my_orders.append("<td>" + data[i].order.name + "</td>");
    my_orders.append("<td>" + data[i].order.invoice + "</td>");
    my_orders.append("<td>" + data[i].order.amount + "</td>");
    my_orders.append("<td>" + data[i].order.response + "</td>");
    my_orders.append("</tr>");
   });   
  });
});

Can you show the backend part? – Gazmend Sahiti Dec 03 '20 at 07:34 — Gazmend Sahiti, Dec 03 '20 at 07:34
I have been updated the question with the backend parts – John Cart Dec 03 '20 at 07:47 — John Cart, Dec 03 '20 at 07:47
have your problem is solved ?? – KUMAR Dec 03 '20 at 10:14 — KUMAR, Dec 03 '20 at 10:14

score 2 · Accepted Answer · answered Dec 03 '20 at 10:47

in the PHP you need to define a $list variable (because you've already used the $data name as an input parameter in your loop), and move the echo outside the loop. Otherwise it echoes each individual data item, rather than creating a coherent JSON array. A list of individual items without the [..] at each end isn't valid JSON, so JavaScript can't read it.

$result = $stmt->fetchAll();
$list = array();

foreach($result as $data => $value)
{
   $list = array('name' => $value["name"], 'amount' => $amount, 'invoice' => $Invoice, 'response' => '200');
}

header("Content-Type: application/json");
echo json_encode($list);
exit();

And, in the JavaScript/jQuery, you have a similar problem where you're using the data name to represent two different things - the list of items, and an individual item within the loop.

This should work better:

$.each(data, function(i, item) {
    my_orders.append("<tr>");
    my_orders.append("<td>" + item.name + "</td>");
    my_orders.append("<td>" + item.invoice + "</td>");
    my_orders.append("<td>" + item.amount + "</td>");
    my_orders.append("<td>" + item.response + "</td>");
    my_orders.append("</tr>");
});

You helped me and I believe you know, Please help me last time, Sir I know you have a big reputation but I am new here. I have limited permissions on statckoverflow. Nobody reply to my question.. Please help me How I append foreach this data in table via Jquery Ajax.. `https://pastebin.com/t1rX3VYX` please pastebin me — John Cart, Dec 07 '20 at 12:17
@JohnCart please ask a new question on StackOverflow about it, so you can provide the code and also a good explanation of the problem, more than fits in a comment. Also, then other people can help you with it too. Paste the link to your new question here and I will take a look if I have some time. — ADyson, Dec 07 '20 at 13:12

Append function() is not working in jQuery Ajax of PHP response

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