Question

我正在尝试使用Retrofit获取和发布数据但它引发异常 Fsaailurecom.google.gson.JsonSyntaxException：java.lang.IllegalStateException：预期BEGIN_OBJECT但在第1行第1行STRING 我尝试了StackOverflow社区建议的所有可能的解决方案，但仍然存在问题退出:( 任何帮助.. !!

PS：在brwoser和Localhost中，我能够获得API call http://onesignaldemo.rf.gd/test/get_user.php的正确Json响应。邮递员返回此网站需要使用Javascript才能运行，请在浏览器中启用Javascript或使用支持Javascript的浏览器

API CALL

private void getUsersFromDB() {
        userservice= ApiUtils.getUserservice();
        userservice.getUsers().enqueue(new Callback<UserResponse>() {
            @Override
            public void onResponse(Call<UserResponse> call, Response<UserResponse> response) {
                Log.d("RESULT", "onResponse: --------------------------------SUCCESS");
                List<Result> userresponses = response.body().getResult();
                Log.d("ANS", "Number of movies received: " + userresponses.size());

            }

            @Override
            public void onFailure(Call<UserResponse> call, Throwable t) {
                Log.d("RESULT", "onResponse: --------------------------------Failure" +t);

            }
        });


    }

UserModal.java

public class UserModel {

    @SerializedName("id")
    @Expose
    private String id;
    @SerializedName("user_name")
    @Expose
    private String userName;
    @SerializedName("user_password")
    @Expose
    private String userPassword;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getUserName() {
        return userName;
    }

    public void setUserName(String userName) {
        this.userName = userName;
    }

    public String getUserPassword() {
        return userPassword;
    }

    public void setUserPassword(String userPassword) {
        this.userPassword = userPassword;
    }

}

UserResponse.java

public class UserResponse {

    @SerializedName("result")
    @Expose
    private List<Result> result = null;
    @SerializedName("error_code")
    @Expose
    private Integer errorCode;

    public List<Result> getResult() {
        return result;
    }

    public void setResult(List<Result> result) {
        this.result = result;
    }

    public Integer getErrorCode() {
        return errorCode;
    }

    public void setErrorCode(Integer errorCode) {
        this.errorCode = errorCode;
    }
}

接口

public interface UserService {
    @GET("/get_user.php")
    Call<UserResponse> getUsers();

    @FormUrlEncoded
    @POST("/insert_user.php")
    Call<DBResponse> insertUsers(@Field("user_name") String username,
                                 @Field("user_password") String userpassword);
}

这是我的Json

{"result":[{"Users":{"id":"1","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"2","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"3","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"4","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"5","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"6","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"7","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"8","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"9","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"10","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"11","user_name":"raj","user_password":"rajesh123"}},{"Users":{"id":"12","user_name":"raj","user_password":"rajesh123"}}]}

Answer 1

服务器只返回一个String，而不是一个对象。试着看看Fiddler的答案是什么，或者只是在Postman中创建一个具有相同请求数据的请求来查看答案。

正如我所看到的，在result中，您有一个对象列表，User中编码的对象会返回您的UserModal.java。因此，您需要在result和UserModal.java之间创建一个中间对象。类似的东西：

public IntermediarUser {
    @SerializedName("User")
    public UserModal user;

    // getter, setter, etc
}

如果您不想这样做，请检查您的服务器实施是否有云。

Answer 2

可能是错的但看起来你的JSON返回一个结果对象，它有一个用户列表而不是结果列表？所以你应该有一个带有List属性的Result pojo，那么你认为你的用户对象应该有三个字符串：id，user_name和user_password。该错误几乎总是（如果不是总是）指的是pojos / data模型结构与JSON结构的不匹配。

预计BEGIN_OBJECT但在第1行STRING - 改造2

2 个答案: