需要一個字符串，但在第 3 行第 4 列路徑 $.SUCCESS 處為 BEGIN_OBJECT

Question

我用 Retrofit2 制作了一個 post 請求方法，但在我的響應中遇到了這個問題。

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

回應應該是

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

我將此代碼用於發布請求

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

對於 POJO 類

public class Post {

    private String anthony;
    private String SUCCESS;

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

    public String getSUCCESS() {
        return SUCCESS;
    }
}

對於我使用以下代碼的方法

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());
            }
        });
    }

有誰知道我的代碼有什么問題？ 我希望在“SUCCESS”json 對象中得到響應。

Answer 1

您希望 SUCCESS 成為您想要的響應中的對象，但您已在 Post 類中將其定義為 String。 你應該使用一個對象來代替 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;
}