Hi,
What is required for the DataHandler is some sort of input soure. So
as long as you have a DataSource that represents the binary then you
are fine!
Here is a code sample for doing that
.... previous code
FileDataSource dataSource = new FileDataSource("yourfile.zip");
expectedDH = new DataHandler(dataSource);
OMText textData = fac.createOMText(expectedDH, true);
....
This should create a proper Datahandler for the OMText to attach the file
Ajith
On 5/4/06, Michele Mazzucco <[EMAIL PROTECTED]> wrote:
Hi Ajith,
thanks for your reply. My questions, actually, is:
In the MTOM sample application, the client has this metho
private OMElement createEnvelope(String fileName) throws Exception {
DataHandler expectedDH;
OMFactory fac = OMAbstractFactory.getOMFactory();
OMNamespace omNs = fac.createOMNamespace("http://localhost/my";,
"my");
OMElement data = fac.createOMElement("mtomSample", omNs);
OMElement image = fac.createOMElement("image", omNs);
Image expectedImage;
expectedImage = new ImageIO()
.loadImage(new FileInputStream(inputFile));
ImageDataSource dataSource = new ImageDataSource("test.jpg",
expectedImage);
expectedDH = new DataHandler(dataSource);
OMText textData = fac.createOMText(expectedDH, true);
image.addChild(textData);
OMElement imageName = fac.createOMElement("fileName", omNs);
if (fileName != null) {
imageName.setText(fileName);
}
//OMElement wrap = fac.createOMElement("wrap",omNs);
// data.addChild(image);
data.addChild(imageName);
data.addChild(image);
//data.addChild(wrap);
return data;
}
If I need to send something different from an image, what function
should I call instead of
Image expectedImage;
expectedImage = new ImageIO().loadImage(new FileInputStream(inputFile));
FileOutputStream imageOutStream = new FileOutputStream(fileName);
new ImageIO().saveImage("image/jpeg", actualObject, imageOutStream);
?
And on the server side, what function should I call instead of
Image actualObject = new ImageIO().loadImage(actualDH.getDataSource()
.getInputStream());
?
Thanks,
Michele
Ajith Ranabahu wrote:
> Hi,
> As far as the content is concerned the zip file is a "binary" file
> just like the image file. All you have to do is read the file through
> a file input stream and then create a new DataSource using that
> inputstream.
>
> Ajith
>
>
> On 5/4/06, Michele Mazzucco <[EMAIL PROTECTED]> wrote:
>> Hi all,
>>
>> the tutorial show how to send images with MTOM, but how can I send
>> binary files with MTOM (say .zip files)?
>>
>> Thanks in advance,
>> Michele
>>
>
>
> --
> Ajith Ranabahu
--
Ajith Ranabahu
