• 05-05-2015, 16:19:14
    #1
    Üyeliği durduruldu
    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
            });
  • 05-05-2015, 18:13:19
    #2
    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.
  • 05-05-2015, 18:20:23
    #3
    kingofseo adlı üyeden alıntı: mesajı görüntüle
    json_decode'nin 2'ci parametre'sini true yapmaniz lazim.
    true veya 1 yapmak aynı şey.