图像数据无法正确发送-Ajax，科尔多瓦

以下代码是使用ajax拍照并发送到服务器的，但是图像数据无法正确发送。

 <script>
      var pictureSource;   // picture source
      var destinationType; // sets the format of returned value 
      var image = "";


  document.addEventListener("deviceready", onDeviceReady, false);

        function onDeviceReady() {

        function alertDismissed() {
        };
            pictureSource = navigator.camera.PictureSourceType;
            destinationType = navigator.camera.DestinationType;
        }

 function capturePhoto() {
            navigator.camera.getPicture(onPhotoDataSuccess, onFail, {
                quality: 50,
                destinationType: destinationType.DATA_URL
            });
        }

            function onPhotoDataSuccess(imageData) {

            var smallImage = document.getElementById('smallImage');

            smallImage.style.display = 'block';

            smallImage.src = "data:image/jpeg;base64," + imageData;

            image = "data:image/jpeg;base64," + imageData;

            alert("Image = "+image);
        }

     function onFail(message) {
            alert('Failed because: ' + message);
        }

    function submitFunction() {

        function alertDismissed() {
        };

        var dataString = 'image='+image;

        $.ajax({
        type: "POST",
        url: "url.php",
        data: dataString,
        cache: false,
        success: function(result){

        }
        });

        }
        </script>

        <input type="button" id="camera" class="btn btn-primary btn-large btn-block" value="Take Photo"  onclick="capturePhoto();"/>

        <input type="submit" class="btn btn-primary btn-large btn-block" value="Next"  onclick="submitFunction();" />

        <img style="display:none;width:60px;height:60px;" id="smallImage" src="" />

图像正确显示，并且警报显示“图像= data：image / jpeg; base64，/ 9j / 4AAQSkZJRgABAQAAAQABAAD / 2wBDABALDA4M ... o7cUAPFJR9KWmgP / 9k =“。

但是，在下面的php中，$ _POST ['image']没有接收到正确的数据，它丢失了一些字符，例如'+'，并且还用空格或换行代替了它。因此，当我从其他页面的db返回时，图像无法正确显示。

url.php

$con=mysql_connect('server','user','password') or die("Failed to connect to MySQL: " . mysql_error());

$db=mysql_select_db('db',$con) or die("Failed to connect to MySQL: " . mysql_error()); 

$retval = mysql_query( "UPDATE tablename SET photo='".$_POST['image']."' WHERE ID='".$_POST['id']."'", $con );

echo $_POST['image'];