Quismo adlı üyeden alıntı: mesajı görüntüle
Ajax ile json tipinde verileri post.php ye gönderiyorum ama verileri çekemiyorum

Warning: Invalid argument supplied for foreach() in

<?php
$array["sonuc"] = json_decode($_POST["data"],1);
foreach ($array["sonuc"] as $key => $value) {
	echo"
	<table width='700' border='0'>
  <tr>
    <td width='284'><b>Product Code</b> : </td>
    <td width='400'>" . $value[0]."</td>
  </tr>
  <tr>
    <td><b>Quantity</b> : </td>
    <td>" .$value[1]."</td>
  </tr>
  <tr>
    <td><b>Price</b> : </td>
    <td>" . $value[2]."</td>
  </tr>
  <!-- <tr>
    <td><b>General Discount</b> : </td>
    <td>" . $value[4]."</td>
  </tr>
  <tr>
    <td><b>SUBTOTAL</b> : </td>
     <td>" . $value[3]."</td>
  </tr> -->
  
</table><br /><br />";



	
}
echo "<h1 style='font-size: 35px;'>SUBTOTAL&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;:<b style='color:#939393'> ". $value[3]."</b></h1>";
echo "<h1 style='font-size: 35px;'>GENERAL TOTAL : <b style='color:#939393'>". $value[2]."</b></h1>";

?>
var dataList = [];
        $('#example tbody').on('click', '#subx', function (sender) {
            debugger
            var data = table.row($(this).parents('tr')).data();

            var dataRow = {
                //sutun0: data[0],
                sutun1: data[1],
                sutun2: data[2],
                sutun3: data[3],
                sutun4: data[4],
                sutun5: data[5]
            };
            dataList.push(dataRow);
            $.ajax({
            type: 'POST',
            datatype:'JSON',
            data:'data='+JSON.stringify(dataList),
            url: 'post.php',
            success: function(result) {
              alert("postedildi");
                $('#dvUrunler').html(result);
            },
        });
            //console.log(JSON.stringify(dataList));

            //debugger
        });
json_decode'nin 2'ci parametre'sini true yapmaniz lazim.