預計BEGIN_OBJECT但STRING在線路1 - 2改造

Question

我試圖獲得並使用改造後的數據，但它拋出異常Fsaailurecom.google.gson.JsonSyntaxException：java.lang.IllegalStateException：預期BEGIN_OBJECT但在行字符串1列1路$ 我試圖通過StackOverflow的社區提出的所有可能的解決方案，但仍是問題退出:( 任何幫助.. !!

PS：在brwoser和本地主機我可以能夠得到正確的JSON響應爲API callhttp://onesignaldemo.rf.gd/test/get_user.php。在郵遞員返回該站點要求Javascript才能正常運行，請在您的瀏覽器上啓用Javascript或者使用瀏覽器的JavaScript支持

API調用

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); 
} 

這就是我的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

可能是錯誤的，但它看起來像你的JSON函數返回一個擁有用戶的列表，而不是結果的列表一個結果對象？所以你應該有一個帶有List屬性的Result pojo，我認爲你的用戶對象應該有三個字符串：id，user_name和user_password。這個錯誤幾乎總是（如果不是總是）指的是你的pojos/data模型的結構和JSON的結構不匹配。

Answer 2

服務器只返回一個字符串，而不是一個對象。試着看看Fiddler的答案是什麼，或者只需在Postman中用相同的請求數據創建一個請求來查看答案。

正如我所見，在result你有一個對象列表，在User編碼的對象，返回您的UserModal.java。因此，您需要在result和UserModal.java之間創建一箇中介對象。例如：

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

    // getter, setter, etc 
} 

如果您不想這樣做，請檢查您的雲服務器實施。