Question

我的json数据看起来像 -

  {
   android: [
               {
                      name: "link1",
                      link: "href="/downloads/-------------""

              },
              {

                   name: "link2",
                   link: "http://www.----------"
               }
           ]
   }

我必须显示仅从我的活动的json数据中获取的名称，以便在显示名称时它应该是可点击的，并且在点击该名称时它应该将其重定向到相应的链接。

**例如：** 当从json数据中提取NAME（link1）并单击时，它应该将其重定向到相应的链接"href="/downloads/-----------"。

我很难让名字可以点击各自的链接。

Answer 1

> coords <-which(mydata!=0, arr.ind=T)
> str(coords)
 int [1:889222, 1:3] 40 40 41 40 41 40 41 40 41 40 ...
 - attr(*, "dimnames")=List of 2
  ..$ : NULL
  ..$ : chr [1:3] "dim1" "dim2" "dim3"

Answer 2

HashMap<String, String> namelink = new HashMap<String, String>();

JsonArray root=response.getJsonArray("android");
for(int i=0;i<root.size();i++)
{
  JsonObject innerElement=root.getJsonObject(i);
  String name=innerElement.getString("name");
  String link=innerElement.getString("link");
  namelink .put(name, link);
}

现在..the hashmap＆＃34; namelink＆＃34;，包含带链接的名称...根据名称获取hashmap值（链接）

例如;

String linkretreived=namelinke.get(<key>);

Answer 3

参考：How to parse JSON in Android首先需要一个JSONParser类：

public class JSONParser {

    static InputStream is = null;
    static JSONObject jObj = null;
    static String json = "";

    // constructor
    public JSONParser() {}

    public JSONObject getJSONFromUrl(String url) {

        // Making HTTP request
        try {
            // defaultHttpClient
            DefaultHttpClient httpClient = new DefaultHttpClient();
            HttpPost httpPost = new HttpPost(url);

            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            is = httpEntity.getContent();

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        try {
            BufferedReader reader = new BufferedReader(new InputStreamReader(
                    is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            is.close();
            json = sb.toString();
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
        }

        // try parse the string to a JSON object
        try {
            jObj = new JSONObject(json);
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        }

        // return JSON String
        return jObj;

    }
}

然后声明活动中所需的所有变量：

// url发出请求 private static String url =“http://www.yourjsonurl.com/json.php”;

// JSON节点名称 private static final String TAG_CONTACTS =“contacts”; private static final String TAG_ID =“id”; private static final String TAG_NAME =“name”;

//联系JSONArray JSONArray contacts = null;

然后：

// Creating JSON Parser instance
JSONParser jParser = new JSONParser();

// getting JSON string from URL
JSONObject json = jParser.getJSONFromUrl(url);

try {
    // Getting Array of Contacts
    contacts = json.getJSONArray(TAG_CONTACTS);

    // looping through All Contacts
    for(int i = 0; i < contacts.length(); i++){
        JSONObject c = contacts.getJSONObject(i);

        // Storing each json item in variable
        String id = c.getString(TAG_ID);
        String name = c.getString(TAG_NAME);
        String email = c.getString(TAG_EMAIL);


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

在android中读取并解析json数据并显示它

3 个答案: