unable to unpack php array with $.post()

Question

I have an array stored in php: $cv1 = array('a','b');

As you can see in the code below, I am trying to get the respective data from the array then delegate it to two separate functions.

But the data returned from the php callback function is: 'Array' instead of 'a','b'; and result[0] gets 'A' result[1] gets 'r' etc.

Thanks for help!

js:

$('a').on('click',function(){
  var cv = $(this).data('cv');
  var url= '_php/myphp.php';
  $.post(url,{contentVar:cv},function(data) {
    result=data;
    return result;
  }).done(function() {
    alert(result[0]);
    $('#myDiv').html(result[1]);
  });
});

php:

$cv1 = array("a","b");

$contentVar = $_POST['contentVar'];

if($contentVar == "cv1")
{
    echo json_encode($cv1);
}

score 3 · Accepted Answer · answered Aug 13 '13 at 16:40

It's common in PHP for you to get "Array" instead of an actual array when it accidentally gets cast to a string. That's not what you're doing here though; we can test:

> $cv1 = array("a","b");
array(2) {
  [0] =>
  string(1) "a"
  [1] =>
  string(1) "b"
}
> json_encode($cv1);
string(9) "["a","b"]"

$cv1 gets encoded correctly.

You must be treating it like a string somewhere else:

> (string)array("a","b")
! Array to string conversion
string(5) "Array"

Also, why do you have two success callbacks in $.post? The 3rd argument is a success callback, which works the same as .done. Use one or the other (I recommend done) but not both.

You may also consider passing json as the last argument (dataType) so that jQuery knows what to expect and will properly decode the result.

Thanks! passing the `json` argument does the job. – BryanK Aug 13 '13 at 21:02 — BryanK, Aug 13 '13 at 21:02

score 0 · Answer 2 · answered Aug 13 '13 at 16:39

0

Try to do something like this:

$.post(url,{contentVar:cv}
   function(response)
   {
      alert(response[0]);
   },'json'
);

answered Aug 13 '13 at 16:39

fagace

116
3

You need to indicate which data type "in this case json" you will use. – fagace Aug 13 '13 at 16:50

score 0 · Answer 3 · edited May 23 '17 at 10:31

You need to convert the json string format to a javascript object/array.

var result = $.parseJSON(data);

see this question jQuery ajax request with json response, how to? for more detail.

But there is something strange that, if is returning 'Array', the php is not converting the array to json, what happens if the ($contentVar == "cv1") return false? If you will return an array to javascript you need to convert to a string (and the json format is perfect for this).

unable to unpack php array with $.post()

3 Answers3