Question

我有一个没有标题的JSON数组。

我一直在努力阅读JSON，但它似乎一直在失败我希望有一种解析JSON而不调用数组名称的例子。

任何帮助或指向我的例子将非常感激。我将在下面附上我遇到错误的代码。这两个网址是我试图阅读的数据。

https://www.descartes.com/rest/glossary-items
https://www.descartes.com/rest/glossary-sources

JsonParser：

public class JsonParser {

    static InputStream is = null;
    static JSONArray jarray = null;
    static String json = "";

    public JSONArray getJSONFromUrl(String url1) {

        StringBuilder builder = new StringBuilder();
        HttpClient client = new DefaultHttpClient();
        HttpGet httpGet = new HttpGet(url1);
        try {
            HttpResponse response = client.execute(httpGet);
            StatusLine statusLine = response.getStatusLine();
            int statusCode = statusLine.getStatusCode();
            if (statusCode == 200) {
                HttpEntity entity = response.getEntity();
                InputStream content = entity.getContent();
                BufferedReader reader = new BufferedReader(
                        new InputStreamReader(content));
                String line;
                while ((line = reader.readLine()) != null) {
                    builder.append(line);
                }
            } else {
                Log.e("==>", "Failed to download file");
            }
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        // try parse the string to a JSON object
        try {
            jarray = new JSONArray(builder.toString());
            // System.out.println(""+jarray);
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        }

        // return JSON String
        return jarray;

    }
}

doInBackground：

protected Void doInBackground(Void... arg0) {
    String str = "";
    JsonParser sh = new JsonParser();

    // Making a request to url and getting response
    String jsonStr = sh.getJSONFromUrl(url1, JsonParser.GET);

    if (jsonStr != null){
        try{
            JSONArray jsonArr = new JSONArray(jsonArr);
            test = jsonArr.getJSONObje(str);
            // looping through All Contacts
            for (int i = 0; i <= str.length(); i++) {

                JSONObject c = str.getJSONObject(i);

                String tid = c.getString(TAG_TID);
                String title = c.getString(TAG_TITLE);
                String acronym = c.getString(TAG_ACRONYM);
                String description = c.getString(TAG_DESCRIPTION);


                // tmp hashmap for single contact
                HashMap<String, String> contact = new HashMap<String, String>();

                // adding each child node to HashMap key => value
                contact.put(TAG_TID, tid);
                contact.put(TAG_TITLE, title);
                contact.put(TAG_ACRONYM, acronym);
                contact.put(TAG_DESCRIPTION, description);

                // adding contact to contact list
                glossaryList.add(contact);
            }
        } catch (JSONException e) {
            e.printStackTrace();
        }
    } else {
        Log.e("JsonParser", "Couldn't get any data from the url");
    }
    return null;
}

Answer 1

首先，您应该在AndroidManifest.xml

中加上这一行

<uses-permission android:name="android.permission.INTERNET"></uses-permission>

2-nd：你的代码中有拼写错误，例如：test = jsonArr.getJSONObje(str); JsonArray中没有这样的方法

3-d：你出于某种原因迭代字符串：for (int i = 0; i <= str.length(); i++) {

所以这里是可以根据您的需求升级的演示代码：

public JSONArray getJSONFromUrl(String url1) {

    String responseBody = "";
    HttpClient client = new DefaultHttpClient();
    HttpGet httpGet = new HttpGet(url1);
    try {
        responseBody = client.execute(httpGet, new BasicResponseHandler());
    } catch (IOException e) {
        e.printStackTrace();
    }
    try {
        JSONArray jarray = new JSONArray(responseBody);
        return jarray;
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }

    return null;

}

protected Void doInBackground(Void... arg0) {
    JsonParser sh = new JsonParser();
    JSONArray jsonArr = sh.getJSONFromUrl("https://www.descartes.com/rest/glossary-items");

    // looping through All Contacts
    if (jsonArr!= null) {
        for (int i = 0; i < jsonArr.length(); i++) {
            JSONObject contact =  jsonArr.optJSONObject(i);
            if (contact!= null)
                Log.e("jsonArr " + i + ":", contact.toString());
        }
    }
    return null;
}

json阅读没有标题的问题数组

1 个答案: