jquery ajax not passing data

Question

On the front-side i am recording user selections:

 var type = $.cookie('liste-voyage-type');  
 var code=  $.cookie('liste-voyage-code');

i then pass these variables to the server

$.ajax({
   url: '../listing-voyage-produit.php', 
   type: "GET",     
   data :  ({type: type,code :code}),
  success: function(data){
         alert('data:'+data);
  }
});

i want to dynamically modify the variables

  $type and $code

at the back end which are used for a sql request that display products on the page

i miss some elements of understanding:

  alert('type:'+type);    => type:2

  alert('code'+code);    => code:Z3\_P95\_

  the success: alert('data:'+data);    =>  data:

the callback function is triggered but the alert(data) does not display anything

how can i play with datatype, json, or eval() to make it working , converting the string variales var type and code to json and ultimately modify the variables on the server side to :

  echo $type;

  echo $code;

if you are not retrieving any data in success, you wont be able to see any data... make sure your success callback return some data... — patel.milanb
– patel.milanb, Commented Dec 13, 2013 at 11:39
what about server side code? What did you do there to get the result? — Felipe Oriani
– Felipe Oriani, Commented Dec 13, 2013 at 11:42
in this function function(data){ data will be the response from server side listing-voyage-produit.php. — Arthur Halma
– Arthur Halma, Commented Dec 13, 2013 at 11:43
does the javascript give you an error upon .get() request and what exactly are you trying to do with the backend? — C. S.
– C. S., Commented Dec 13, 2013 at 11:45

Mr.G · Accepted Answer · 2013-12-13 11:54:17Z

2

your code missing dataType

$.ajax({
   url: '../listing-voyage-produit.php', 
   type: "GET",     
   data :  {type:type,code:code},
   dataType:"json",
  success: function(data){
         alert('data:'+data);
  }
});

answered Dec 13, 2013 at 11:54

Mr.G

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Comments

C. S. · Accepted Answer · 2013-12-13 12:06:29Z

1

Do not wrap your data with parenthesis ()

JAVASCRIPT

...
data :  {type: type,code :code},
dataType: 'json',
...

PHP

// decode data sent
$json = json_decode($_POST['data']);

// encode data to send back
echo json_encode($json);

edited Dec 13, 2013 at 12:06

answered Dec 13, 2013 at 11:45

C. S.

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2 Comments

A. Wolff Over a year ago

this won't solve OP issue but that's right than here parenthesis are useless (but still valid)

C. S. Over a year ago

what are you trying to do exactly? pass the data to a php file and return the same exact data back?

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