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Trying to upload a file to a JAX-RS (jersey) server

Asked 2011-04-24T17:56:19.087
14

I'm trying to upload a file and other form data using multipart/form-data client with Jersey. I'm uploading to a REST web service also using Jersey. Here is the server code:

@POST
@Consumes(MediaType.MULTIPART_FORM_DATA)
@Produces(MediaType.APPLICATION_JSON)
public String create(@FormDataParam("file") InputStream file,
        @FormDataParam("file") FormDataContentDisposition fileInfo,
        @FormDataParam("name") String name,
        @FormDataParam("description") String description) {
    Ingredient ingredient = new Ingredient();
    ingredient.setName(name);
    ingredient.setDescription(description);
    ingredient.setImageName(fileInfo.getFileName());
    ingredient.setImagePath(context.getRealPath("/resources/uploads/"));
    // TODO save the file.
    try {
        JSONObject json = new JSONObject();
        try {
            ingredientService.create(ingredient);
        } catch (final InvalidParameterException ex) {
            logger.log(Level.INFO, ex.getMessage());
            json.put("result", false);
            json.put("error", ex.getMessage());
            return json.toString();
        } catch (final GoodDrinksException ex) {
            logger.log(Level.WARNING, null, ex);
            json.put("result", false);
            json.put("error", ex.getMessage());
            return json.toString();
        }
        json.put("ingredient", JsonUtil.ingredientToJSON(ingredient));
        return json.put("result", true).toString();
    } catch (JSONException ex) {
        logger.log(Level.SEVERE, null, ex);
        return "{\"result\",false}";
    }
}

I've tested the server code using a basic html form on my desktop and it works fine. The problem seems to be in the client. Here is the relevant client code.

ClientConfig config = new DefaultClientConfig();
client = Client.create(config);
client.addFilter(new LoggingFilter());
webResource = client.resource("http://localhost:8080/webapp/resources").pa
java file-upload jersey client jax-rs
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1 Answer

-2

Or just write a new file and upload it: 

Writer output = null;
    File file = null;
    try {
      String text = "Rajesh Kumar";
      file = new File("write.txt");
      output = new BufferedWriter(new FileWriter(file));
        output.write(text);
        output.close();
    } catch (IOException e) {
        System.out.println("IOException e");
        e.printStackTrace();
    }

    InputStream is = null;

    try {
        is = new FileInputStream(file);
    } catch (FileNotFoundException e) {
        System.out.println("FileNotFoundException e");
        e.printStackTrace();
    } catch (IOException e) {
        System.out.println("IOException e");
        e.printStackTrace();
    }

    FormDataMultiPart part = new FormDataMultiPart().field("file", is, MediaType.TEXT_PLAIN_TYPE);
    res = service.path("rest").path("tenant").path(tenant1.getTenantId()).path("file").type(MediaType.MULTIPART_FORM_DATA_TYPE).post(ClientResponse.class, part);
answered 2012-06-21T15:54:40.020

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