So confused about how to connect frontend and backend?

Suppose I've got an object obj and var jsonText = JSON.stringify(obj);

how can I send this jsonText to backend (locally, non remote server, no database), using php to get this data and then save the content as a single new JSON file?

Thanks a lot!

So confused about how to connect frontend and backend?

Suppose I've got an object obj and var jsonText = JSON.stringify(obj);

how can I send this jsonText to backend (locally, non remote server, no database), using php to get this data and then save the content as a single new JSON file?

Thanks a lot!

• javascript
• php
• connection
• frontend
• backend

Share Follow asked Dec 30, 2015 at 23:07 KAFFEECKOKAFFEECKO 17311 gold badge22 silver badges1010 bronze badges 9

• Ajax. – Script47 Commented Dec 30, 2015 at 23:07
• @Script47 but how to define the post target address? – KAFFEECKO Commented Dec 30, 2015 at 23:10
• Using the url parameter which is available. – Script47 Commented Dec 30, 2015 at 23:11
• Check this page to see how to run local php application. – Ahamed Commented Dec 30, 2015 at 23:13
• 1 it would be: /your-file.php which is located on http://localhost/ .. the full path might be something like http://localhost/your-file.php and as @Ahamed posted, make sure you have a web server running otherwise what I said is not possible. – Clay Commented Dec 30, 2015 at 23:14

 |  Show 4 more ments

1 Answer 1

Sorted by: Reset to default 3

You need to query your data from a database or from somewhere else in PHP. Then, you can echo it with PHP in a JSON format. In a second step, you can use jQuery or plain JavaScript to make an Ajax call to this PHP file and make something with it.

PHP (data.json.php):

<?php
  header('Content-Type: application/json');
  $output = array();
  // query the data from somewhere
  $output['data'] = "1234";
  echo json_encode($output); // json_encode creates the json-specific formatting
?>

JavaScript (jQuery):

$.ajax({
   url: "data.json.php",
   success: function(result){
      console.log(result);
   }
});

The code is untested, but I think you should get the idea.