get data from php using ajax

Question

I have the following ajax function:

var data=event.target.result;
var fileName=encodeURIComponent('audio_recording_' + new Date().getMinutes() + '.wav');
$.ajax({
    type: 'POST',
    url: 'readFile.php',
    data: {"fileName":fileName,"data":data},
    success: function(data){
        console.log(data);
    }
});

the server-side code

<?php
$fileName=$_POST["fileName"];
$data=$_POST["data"];
$dh = opendir('upload/');
$contents = file_get_contents('C:/wamp/www/JSSoundRecorder/upload/'.$fileName);
// echo $contents;
echo $fileName;
echo $data;

In the console.log(data) i'm obtaining the correct results (filename and data) what I want is to have each info alone to use later. that is fileName in a variable and data in another variable in the success function so that I can use them later in the program.I think I should use maybe localstorage and json.stringify??is that correct.or is there another way.and if that is true can you help me how to use localstorage here? thank you in advance

Think Different · Accepted Answer · 2014-07-17 07:52:43Z

Try this

var data=event.target.result;
var fileName=encodeURIComponent('audio_recording_' + new Date().getMinutes() + '.wav');
$.ajax({
   type: 'POST',
   url: 'readFile.php',
   data: {"fileName":fileName,"data":data},
   dataType: 'json', //<==== Add this
   success: function(data){
     console.log(data.filename);
     console.log(data.data);
   }

});

and you php should be:

<?php
   $fileName=$_POST["fileName"];
   $data=$_POST["data"];
   $dh = opendir('upload/');
   $contents = file_get_contents('C:/wamp/www/JSSoundRecorder/upload/'.$fileName);
   $data = array('filename'=>$filename, 'data'=>$data); 
   echo json_encode($data);
?>

dhalfageme · Accepted Answer · 2014-07-17 07:59:12Z

0

You can put the info in an object or an associative array an then use json_encode, that converts the object in a serialized string. Then you can parse the JSON string in Javascript:

In php, if you have an object or assocciative array, you will write:

echo json_encode($yourObjectOrArray);

Supose data is the data you receive in the success function:

var data = '{"fileName":"yourFileName","data": "yourData", "moreData" : "moreData"}';
var obj = JSON.parse(json);

edited Jul 17, 2014 at 7:59

answered Jul 17, 2014 at 7:50

dhalfageme

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Comments

Dwza · Accepted Answer · 2014-07-17 07:51:11Z

0

try

php

$result = array("filename" => $fileName, "data" => $data);
echo json_encode($result);

jquery

success: function(response){
    console.log(response);
    // var myObject = $.parseJSON(response)
    // here you have an object with all values now
}

myObject is needed, depending on if you use dataType: "JSON" in your ajax or not.

answered Jul 17, 2014 at 7:51

Dwza

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Comments

Yves Lange · Accepted Answer · 2014-07-17 07:52:29Z

You should use JSON in javascript part to have it as an object.

You can use var jsonData = JSON.parse(data.data)

On the server you should encode it in json with the corresponding PHP function: json_encode(YOUR_ARRAY)

More information here: https://developer.mozilla.org/fr/docs/Web/JavaScript/Reference/Objets_globaux/JSON/parse

There is a great example here: http://www.caveofprogramming.com/frontpage/articles/php/php-json-an-example-javascript-json-client-with-php-server/

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