需要一个字符串，但在第 3 行第 4 列路径 $.SUCCESS 处为 BEGIN_OBJECT

Question

我用 Retrofit2 制作了一个 post 请求方法，但在我的响应中遇到了这个问题。

Expected a string but was BEGIN_OBJECT at line 3 column 4 path $.SUCCESS

回应应该是

{
"SUCCESS" :
{
"200" : "access granted",
"ra" : "approved",
"la" : "approved",
"ch" : "approved"
}
}

我将此代码用于发布请求

@POST("login")
Call<Post> createPost(@Body Post post);

对于 POJO 类

public class Post {

    private String anthony;
    private String SUCCESS;

    public Post(String name) {
        this.anthony = name;
    }

    public String getSUCCESS() {
        return SUCCESS;
    }
}

对于我使用以下代码的方法

private void createPost() {
        Post post = new Post("mypassword");
        Call<Post> call = jsonPlaceHolderApi.createPost(post);

        call.enqueue(new Callback<Post>() {
            @Override
            public void onResponse(Call<Post> call, Response<Post> response) {
                if (!response.isSuccessful()) {
                    textViewResult.setText("Code: " + response.code());
                    return;
                }
                Post postResponse = response.body();
                String content = "";
                content += "Code: " + response.code() + "\n";
                content += "S" + postResponse.getSUCCESS();
                textViewResult.setText(content);
            }
            @Override
            public void onFailure(Call<Post> call, Throwable t) {
                textViewResult.setText(t.getMessage());
            }
        });
    }

有谁知道我的代码有什么问题？ 我希望在“SUCCESS”json 对象中得到响应。

Answer 1

您希望 SUCCESS 成为您想要的响应中的对象，但您已在 Post 类中将其定义为 String。 你应该使用一个对象来代替 SUCCESS。

public class Post {

    private String anthony;
    private PostSuccess SUCCESS;

    public Post(String name) {
        this.anthony = name;
    }

    public PostSuccess getSUCCESS() {
        return SUCCESS;
    }
}

public class PostSuccess {
    @JsonProperty("200")
    private String _200;
    private String ra;
    private String la;
    private String ch;
}