php array not forming properly

Question

Initial javascript to get a string/array:

<script type="text/javascript">
$(document).ready(function() {
    $("#table").tableDnD({
    onDrop: function(table, row){
        var rows = table.tBodies[0].rows;
            var debugStr = "";
            for (var i=1; i<rows.length; i++) {
                debugStr += rows[i].id+":";
                //debugStr += rows[i].id;
            }
        data = debugStr; // colon seperated
        }
    });
});
function sendData()
{
    var packed = ""; // Initialize packed or we get the word 'undefined'
    for (i = 0; (i < data.length); i++) {
    if (i > 0) {
        packed += ",";
        }
    packed += escape(data[i]);
        }
    document.data.data.value = packed;
    document.data.submit();
}
</script>

Now in the PHP...

Return my array/string from post:

$packed = $_POST['data'];
$data = split(":", $packed);
for ($i = 0; ($i < count($data)); $i++) {
  $data[$i] = rawurldecode($data[$i]);
  $data[$i] = str_replace(",", "", $data[$i]);
  $data[$i] = str_replace(":", ",", $data[$i]);
}

create an array from it:

$arrayUpdated = array();
foreach($data as $value) {
    print_r($value);
    $arrayUpdated[] = $value;
    }

which returns:

Array ( [0] => 9,8,6,5,4,3,2,15,1,0, )

when I was expecting:

Array ( [0] => 9, [1] => 8, [2] => 6, [3] => 5, [4] => 4, [5] => 3, [6] => 2, [7] => 15, [8] => 1, [9] => 0, )

what have I got wrong? Thanks, Andy

What happens if you do print_r($packed) straight after it intercepts the POST array? — SW4
– SW4, Commented Nov 1, 2010 at 13:35
er... so sendData inserts a comma after each character in data? Why? — Powerlord
– Powerlord, Commented Nov 1, 2010 at 15:37

Spudley · Accepted Answer · 2010-11-01 14:29:12Z

2

Since you're using JQuery already, wouldn't it be easier to send the data in JSON format rather than manually building it into a colon/comma separated and then manually spliting it apart in PHP?

In your JS page, replace most of your sendData() function with a call to $.toJSON(data);:

function sendData() {
    document.data.data.value = $.toJson(data);
    document.data.submit();
}

In your PHP code, simlpy use json_decode() to load your data into an array with exactly the same structure as you had in Javascript.

$data=json_decode($_POST['data']);
print_r($data);

*Note: you may need an extra JQuery plugin to get the toJSON() method -- See http://code.google.com/p/jquery-json/ for more info.

edited Nov 1, 2010 at 14:29

answered Nov 1, 2010 at 13:39

Spudley

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4 Comments

Spudley Over a year ago

edited to include code examples and a link to where you can get the $.toJSON() plug-in.

Andy Over a year ago

nice one, i'm going to have a go at putting this in, it should simplify my code a fair bit. cheers!

Andy Over a year ago

Tried this, nothing showing up... I've got the 'jquery.json-2.2.js' include correctly linked in. I inserted a js alert to check contents of 'data' (its all in there) then when the posted page is returned and I do: $data=json_decode($_POST['data']); print_r($data); nothing shows, any ideas why?

Spudley Over a year ago

Okay, so the JQuery JSON encode side is working. That's good. Have you checked the contents of $_POST before doing json_decode()? If that's blank then it's presumably the form that isn't posting correctly. I took the document.data.data.value and document.data.submit directly from your original code, so its possible they may not be correct. Try checking them in Firebug (or a similar browser debugging tool).

SW4 · Accepted Answer · 2010-11-01 13:32:16Z

1

Not a direct answer- but if you need a quick fix you could do (in PHP):

// if $myarray=Array ( [0] => 9,8,6,5,4,3,2,15,1,0, ) ;

$myarray=explode(",",$myarray[0]);

Also, note that the split function is deprecated in PHP 5.3.0, you can replace with explode.

answered Nov 1, 2010 at 13:32

SW4

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1 Comment

Andy Over a year ago

This did the trick for how I had it, but ended up using JSON as above instead.

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