引起：java.lang.NullPointerException：尝试在空对象引用上调用虚方法'int org.json.JSONArray.length（）'[duplicate]

Posted 2021-05-02

这个问题在这里已有答案：

• What is a NullPointerException, and how do I fix it? 12个答案

我需要从json解析String builder，但我收到此错误：

Caused by: java.lang.NullPointerException: Attempt to invoke virtual method 'int org.json.JSONArray.length()' on a null object reference

我不知道为什么stringBuilder.toString()给我带有物体的数组！

这是我的代码：

try {
    // HttpResponse is an interface just like HttpPost.
    //Therefore we can't initialize them
    HttpResponse httpResponse = httpClient.execute(httpPost);

    // According to the JAVA API, InputStream constructor do nothing.
    //So we can't initialize InputStream although it is not an interface
    InputStream inputStream = httpResponse.getEntity().getContent();

    InputStreamReader inputStreamReader = new InputStreamReader(inputStream);

    BufferedReader bufferedReader = new BufferedReader(inputStreamReader);

    StringBuilder stringBuilder = new StringBuilder();

    String bufferedStrChunk = null;

    while((bufferedStrChunk = bufferedReader.readLine()) != null){
        stringBuilder.append(bufferedStrChunk);
    }

    // return stringBuilder.toString();
    System.out.println("EHEH"+stringBuilder.toString());


    JSONArray mainObject = null;
    try {
        mainObject = new JSONArray(stringBuilder.toString());
    } catch (JSONException e) {
        e.printStackTrace();
    }

    for (int i = 0; i < mainObject.length(); i++) {
        JSONObject object = null;
        try {
            object = mainObject.getJSONObject(i);
        } catch (JSONException e) {
            e.printStackTrace();
        }
        try {
            Video video=new Video();
            String url = object.getString("url");

            String titolo = object.getString("titolo");
            String sottotitolo = object.getString("sottotitolo");
            String data = object.getString("date");
            //video.setId(idd4);
            System.out.println("CAZZO:"+video.getId());
            video.setPic(url);
            video.setTitolo(titolo);
            video.setSottotitolo(sottotitolo);
            video.setData(data);

            videoList.add(video);
        } catch (JSONException e) {
            e.printStackTrace();
        }
    }

有人能帮我吗？

JSON响应：

答案

你的错误是String data = object.getString("data");

JSONObject mainObject = null;
    try 
    {
        mainObject = new JSONObject(stringBuilder.toString());
        JSONArray jsonArrayPDF = mainObject.getJSONArray("pdfs");

        for (int i = 0; i < jsonArrayPDF.length(); i++) {

        JSONObject object = jsonArrayPDF.getJSONObject(i);
        Video video=new Video();
        String url = object.getString("url");

        String titolo = object.getString("titolo");
        String sottotitolo = object.getString("sottotitolo");
        String data = object.getString("data");
        //video.setId(idd4);
        System.out.println("CAZZO:"+video.getId());
        video.setPic(url);
        video.setTitolo(titolo);
        video.setSottotitolo(sottotitolo);
        video.setData(data);

        videoList.add(video);

    }

    } 
    catch (JSONException e) 
   {
        e.printStackTrace();
    }

另一答案

stringBuilder.toString()回归了什么？使用System.out.println()

另一答案

你忘了初始化StringBuilder。

首先初始化然后再使用

StringBuilder sb = new StringBuilder();