未定义的变量使用Ajax之后

Question

I am doing a small web application. The purpose of this application is storing data to my database. Here is my application process:

First ,the user will upload they file to my web. Here is how file format looks like.

The test.txt will record that each companys sell different items on the date.

test.txt

companyA
date price name
02/12 9.99 name1
...   ... ...
12/01 996.00 name100
companyB
date price name
02/12 9.99 name2
...   ...
12/01 956.00 name200
companyC
date price name
02/12 4.99 name5
...   ...
12/01 996.00 name200
companyD
date price name
02/12 59.99 name1
...   ...
12/01 96.00 name400

Second step: Once use uploaded test.txt. I will display companyA,companyB,companyC and companyD to front-end.(I use <div id="company_result0" style="color:#0A0A0A"></div>).When I display company ,i use radio button for each company name.

    $('#company_result0').append($('<input type="radio" name="companyName" value="' + (key+1) + '" > ' + value + '</input><br />'));

Third step: Let use choose which company information they want to store.

index.html

<form name="uploadform" id="uploadform"  >
<input id="file_browse" type="file" class="btn btn-default text-center defbtntext"  name="fileToUpload">
 </form>
<form class="data-display" id="data-display" name ="data-display">
<div id="company_result0"  style="color:#0A0A0A"></div>
</form>

ajax.js

function getUploadFile() {
        $('#uploadform').on('change', function(e) {
              e.preventDefault();
              var formData = new FormData($(this)[0]);
              $.ajax({
                  type : "POST",
                  url : "findName.php",
                  data : formData,
                  processData: false,
                  contentType: false,
                  success: function (result) {

            var entry = JSON.parse(result);
            $.each(entry ,function(key,value) {
            $('#company_result0').append((key+1));
            $('#company_result0').append(": ");
            $('#company_result0').append($('<input type="radio" name="companyName" value="' + (key+1) + '" > ' + value + '</input><br />'));


        });
 ('#company_result0').append($('<input type="submit" id="submitCompanyName" name="submitCompanyName"</input><br />'));
                 }

              });
        });
}

findName.php

$tempfile = $_FILES ['fileToUpload'] ['tmp_name'];
$fileInfo = new getFileInfo ( $tempfile );
$fileInfo->findcompanyname ();
$fileInfo->sendcompanynametofront ();//send each company name to front.

/**company names will display in #company_result0.
*How can I to get value from #company_result0?
*Here is my code for get value, but it is not work.
*it will show me undefined variable :companyName error.
*/

if ($_SERVER ["REQUEST_METHOD"] == "POST") {
    if (isset($_POST['companyName'])) {
        $companyName=$_POST['companyName'];

    }
}

getFileInfo.php

function sendcompanynametofront() {
    echo json_encode ( companyNameArray );
}

My question is How can I to get value from #company_result0.

When I click submit button in #company_result0 ,it will give the error which is undefined variable :companyName.

Please let me know if there are any misunderstanding.