CODEWITHSUNDEEP
javascriptphpajax
lama
lama

Reputation: 129

get data from php using ajax

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

Upvotes: 0

Views: 74

Answers (4)

dhalfageme
dhalfageme

Reputation: 1545

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);

Upvotes: 0

Think Different
Think Different

Reputation: 2815

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);
?>

Upvotes: 1

Yves Lange
Yves Lange

Reputation: 3994

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/

Upvotes: 0

Dwza
Dwza

Reputation: 6565

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.

Upvotes: 0

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