改造后期参数

Question

我正在实现登录功能,并使用Post请求,但我收到错误说

"retrofit.RetrofitError:com.squareup.okhttp.internal.http.HttpMethod.METHODS"

以下是我的代码

import java.util.HashMap;
import java.util.Map;

import retrofit.Callback;
import retrofit.http.*;




//Myapi.java

import java.util.HashMap;
import java.util.Map;

import retrofit.Callback;
import retrofit.http.*;

public interface MyApi {

    /* LOGIN */
    @POST("/api/0.01/oauth2/access_token/")
    // your login function in your api
    public void login(@Body HashMap<String, String> arguments, Callback<String> calback);
}


//In my activity
RestAdapter restAdapter = new RestAdapter.Builder()
                .setEndpoint(Constants_Interface.URL).setClient(newclient)
                .build();

        MyApi mylogin = restAdapter.create(MyApi.class); 
HashMap<String, String> dicMap = new HashMap<String, String>();
dicMap.put("client_id", XXX);
        dicMap.put("client_secret", XXX);
        dicMap.put("username", XXX);
        dicMap.put("password", XXX);
mylogin.login(dicMap, new Callback<String>() {

            @Override
            public void failure(RetrofitError retrofitError) {
                retrofitError.printStackTrace(); // to see if you have
                                                    // errors
            }

            @Override
            public void success(String s, retrofit.client.Response response) {
                // TODO Auto-generated method stub
                Toast.makeText(getApplicationContext(), "Login Succes",
                        Toast.LENGTH_LONG).show();

            }
        });

下面是logcat输出.

02-10 13:02:43.846:W/System.err(30684):retrofit.RetrofitError:com.squareup.okhttp.internal.http.HttpMethod.METHODS 02-10

Answer 1

试试这个

public interface SafeUserApi {
 @FormUrlEncoded
    @POST("/api/userlogin")
    void getUserLogin(
            @Field("client_id") String id,
            @Field("client_secret") String secret,
            @Field("username") String uname,
            @Field("password") String password,
            Callback<LoginResult> cb
    );
}

这里parm1是POST参数,您将把它传递给服务器.这将解决您的问题

如果你使用PHP,你可以使用访问param1 $uname= $_POST('username');

编辑1:

改造2.0版本:

public interface SafeUserApi {
    @FormUrlEncoded
    @POST("/api/userlogin")
    Call<ResponseBody>  getUserLogin(
            @Field("client_id") String id,
            @Field("client_secret") String secret,
            @Field("username") String uname,
            @Field("password") String password
    );
}

Answer 2

您还可以传递多个字段参数: 例如:

@FormUrlEncoded
@POST("/oauth/access_token")
void getToken(
    @FieldMap Map<String, String> params, 
    Callback<FacebookLoginUserResponse> callback
);

Answer 3

改造2.0版本：

@FormUrlEncoded
@POST("api/v2/users/sign_in")
Call<SignInResult> userSignIn(
        @FieldMap HashMap<String, String> authData
);