如何从php服务器获取json数据到android手机

Question

我有一个应用程序,我想从php web服务器获取json数据到android手机.我所拥有的是一个网址并点击该网址给我json数据
{"items":[{"latitude":"420","longitude":"421"}]}.但我想在我的Android手机中检索这个json格式,并从json格式获取纬度和经度的值.

我们怎么能在Android手机上得到它？

提前致谢..

Answer 1

首先进行URL连接

String parsedString = "";

    try {

        URL url = new URL(yourURL);
        URLConnection conn = url.openConnection();

        HttpURLConnection httpConn = (HttpURLConnection) conn;
        httpConn.setAllowUserInteraction(false);
        httpConn.setInstanceFollowRedirects(true);
        httpConn.setRequestMethod("GET");
        httpConn.connect();

        InputStream is = httpConn.getInputStream();
        parsedString = convertinputStreamToString(is);

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

JSON字符串

{
"result": "success",
"countryCodeList":
[
  {"countryCode":"00","countryName":"World Wide"},
  {"countryCode":"kr","countryName":"Korea"}
] 
}

我在下面提到国家详细信息

JSONObject json = new JSONObject(jsonstring);
JSONArray nameArray = json.names();
JSONArray valArray = json.toJSONArray(nameArray);

JSONArray valArray1 = valArray.getJSONArray(1);

valArray1.toString().replace("[", "");
valArray1.toString().replace("]", "");

int len = valArray1.length();

for (int i = 0; i < valArray1.length(); i++) {

 Country country = new Country();
 JSONObject arr = valArray1.getJSONObject(i);
 country.setCountryCode(arr.getString("countryCode"));                        
 country.setCountryName(arr.getString("countryName"));
 arrCountries.add(country);
}




public static String convertinputStreamToString(InputStream ists)
        throws IOException {
    if (ists != null) {
        StringBuilder sb = new StringBuilder();
        String line;

        try {
            BufferedReader r1 = new BufferedReader(new InputStreamReader(
                    ists, "UTF-8"));
            while ((line = r1.readLine()) != null) {
                sb.append(line).append("\n");
            }
        } finally {
            ists.close();
        }
        return sb.toString();
    } else {
        return "";
    }
}