java - 使用 @FormParam 的 PUT 方法

Question

如果我有类似的东西：

@PUT
@Path("/login")
@Produces({"application/json", "text/plain"})
@Consumes("application/json")
public String login(@FormParam("login") String login, @FormParam("password") String password) throws Exception
{
    String response = null;
    response = new UserManager().login(login, password);
    return response;
}

如何输入这两个参数来测试我的 REST 服务（在内容字段中）？不是这样的：

{"login":"xxxxx","password":"xxxxx"}

谢谢

Answer 1

Form parameter data will only be present when you submit ... form data. Change the @Consumes type for your resource to multipart/form-data.

@PUT
@Path("/login")
@Produces({ "application/json", "text/plain" })
@Consumes("multipart/form-data")
public String login(@FormParam("login") String login,
        @FormParam("password") String password) {
    String response = null;
    response = new UserManager().login(login, password);
    return response;
}

Then on your client side, set:

• Content-Type: multipart/form-data
• Add form variables for login and password

On a side note, assuming this isn't for learning, you will want to secure your login endpoint with SSL, and hash the password before sending it across the wire.

---

EDIT

Based on your comment, I am including an example of sending a client request with the required form data:

try {
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost post = new HttpPost(BASE_URI + "/services/users/login");

    // Setup form data
    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
    nameValuePairs.add(new BasicNameValuePair("login", "blive1"));
    nameValuePairs.add(new BasicNameValuePair("password",
            "d30a62033c24df68bb091a958a68a169"));
    post.setEntity(new UrlEncodedFormEntity(nameValuePairs));

    // Execute request
    HttpResponse response = httpclient.execute(post);

    // Check response status and read data
    if (response.getStatusLine().getStatusCode() == HttpStatus.SC_OK) {
        String data = EntityUtils.toString(response.getEntity());
    }
} catch (Exception e) {
    System.out.println(e);
}