获取 JSONException？

Question

我试图解析下面的 json 并在第 150 行（在代码中标记）捕获了 JSONException，但我无法弄清楚为什么会这样。 Objective 是找到特定包的测试列表并将其存储在 arraylist 中。

当我通过 screen_package 作为第二个参数时，日志消息是： Value screen_package of type java.lang.String cannot转换为 JSONObject。根据我的理解，我们可以在 JSONObject 的构造函数中传递 String 对象。如果我错了，请纠正我。看看

{
  "package_name": {
    "camera_package": {
      "RearCamera": {
        "test_name": "RearCamera",
        "test_type": "Mandatory",
        "visibility": "true",
        "report_name": "RearCamera",
        "test_category": "manual",
        "order": "42"
      },
      "FrontCamera": {
        "test_name": "FrontCamera",
        "test_type": "Mandatory",
        "visibility": "true",
        "report_name": "FrontCamera",
        "test_category": "manual",
        "order": "43"
      },
      "Flash": {
        "test_name": "Flash",
        "test_type": "Mandatory",
        "visibility": "true",
        "report_name": "Flash",
        "test_category": "manual",
        "order": "1"
      },
      "AutoFocus": {
        "test_name": "AutoFocus",
        "test_type": "Mandatory",
        "visibility": "true",
        "report_name": "AutoFocus",
        "test_category": "manual",
        "order": "35"
      },
      "VideoRecord": {
        "test_name": "VideoRecord",
        "test_type": "mandatory",
        "visibility": "true",
        "report_name": "VideoRecord",
        "test_category": "manual",
        "order": "10"
      },
      "FrontVideoRecorder": {
        "test_name": "FrontVideoRecorder",
        "test_type": "mandatory",
        "visibility": "true",
        "report_name": "FrontVideoRecorder",
        "test_category": "manual",
        "order": "11"
      }
    },
    "screen_package": {
      "Screen": {
        "test_name": "Screen",
        "test_type": "Mandatory",
        "visibility": "true",
        "report_name": "Screen",
        "test_category": "manual",
        "order": "21"
      }
    }
  }
}

以下是 java 代码：

public void parseJSON(String jsonName, String packageName) {
        try {
            JSONObject jsonRootObject = new JSONObject(jsonName);
            if (jsonRootObject != null && jsonRootObject.length() > 0) {
                Iterator<String> packageKeys = jsonRootObject.keys();
                if (packageKeys != null) {
                    while(packageKeys.hasNext()){
                        String currentPackageKey = packageKeys.next();
                        if(currentPackageKey!= null && currentPackageKey.length()>0 &&
                                currentPackageKey.equals(packageName)){
                            JSONObject jsonPackageObject = new JSONObject(currentPackageKey);  //line 150
                            if(jsonPackageObject!=null && jsonPackageObject.length()>0) {
                                Iterator<String> testNameKeys = jsonPackageObject.keys();
                                if(testNameKeys!=null){
                                    while(testNameKeys.hasNext()) {
                                        String currentTestNameKey = testNameKeys.next();
                                        if(currentTestNameKey!=null && currentTestNameKey.length()>0) {
                                            //do something
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }

        } catch (JSONException e) {
            e.printStackTrace();
            System.out.println("Exception caught : "+e.getLocalizedMessage());
        }
    }

Answer 1

在这

JSONObject jsonPackageObject = new JSONObject(currentPackageKey);

currentPackageKey 是一个键 String.You 必须在那里提供 JSON 格式化字符串。

所以用

代替

JSONObject jsonPackageObject = jsonRootObject.getJSONObject(currentPackageKey);

Answer 2

JSONException表示在JSON处理.

期间发生了一些异常

你应该使用 getJSONObject(String name).

JSONObject getJSONObject (String name)

Returns the value mapped by name if it exists and is a JSONObject, or throws otherwise

代码结构

 JSONObject screen_package = jsonobject.getJSONObject("screen_package");

获取 JSONException？

Getting JSONException?

android

json

jsonexception