Question

我尝试使用此网址 www.example.com/directory / 从服务器获取数据，但它无法正常工作......它想出了这个json没有向服务器发送任何动作......所以我想要的就是以这种方式制作网址 - www.example.com/directory/index.php?action=searchFood&money=50

这是我的JsonFetcher.java

public class JsonFetcher extends AsyncTask<Pair<String, String>, Integer, JSONObject> {

private JSONObject jsonObject;
public AsyncResponse delegate = null;

public interface AsyncResponse {
    void processFinish(JSONObject output);
}

public JsonFetcher(AsyncResponse delegate) {
    this.delegate = delegate;
}

@Override
protected void onPostExecute(JSONObject jsonObject) {
    super.onPostExecute(jsonObject);
    delegate.processFinish(jsonObject);
}

@Override
protected JSONObject doInBackground(final Pair<String, String>... params) {

    int count = params.length;
    URL url = null;
    int responseCode = 0;
    jsonObject = null;
    try {
        url = new URL("http://www.example.com/directory/");

    } catch (MalformedURLException e) {
        e.printStackTrace();
    }
    HttpURLConnection conn = null;
    try {
        conn = (HttpURLConnection) url.openConnection();
        //conn.connect();
    } catch (IOException e) {
        e.printStackTrace();
    }
    try {
        conn.setRequestMethod("POST");
        conn.setDoInput(true);
        conn.setDoOutput(true);
        Uri.Builder builder = new Uri.Builder();
        Date date = new Date();
        String md5 = date.toString();

        //Log.e("json lock",lock);
        //Log.e("json key",md5);

        for (int i = 0; i < count; i++) {
            builder.appendQueryParameter(String.valueOf(params[i].first),String.valueOf(params[i].second));
        }
        String query = builder.build().getEncodedQuery();
        OutputStream os = conn.getOutputStream();
        BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));
        writer.write(query);
        writer.flush();
        writer.close();
        os.close();
    } catch (ProtocolException e) {
        e.printStackTrace();
    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    // read the response
    try {
        responseCode = conn.getResponseCode();
    } catch (IOException e) {
        e.printStackTrace();
    }

    if (responseCode == HttpsURLConnection.HTTP_OK) {
        InputStream in = null;
        try {
            in = new BufferedInputStream(conn.getInputStream());
        } catch (IOException e) {
            e.printStackTrace();
        } finally {
            //conn.disconnect();
        }

        BufferedReader r = new BufferedReader(new InputStreamReader(in));
        StringBuilder total = new StringBuilder();
        String line;
        try {
            while ((line = r.readLine()) != null) {
                total.append(line);
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
        Log.e("json", total.toString());
        try {
            jsonObject = new JSONObject(total.toString());
        } catch (JSONException e) {
            e.printStackTrace();
        }
    }

    return jsonObject;
}

}

是否有可能以这种方式写url = new URL("http://www.example/directory/index.php?action= PairActionName & PairMoney" );这里PairActionName和PairMoney将从MainActivity接收

Answer 1

由于您使用AsyncTask执行服务器操作，因此您应创建一个编码查询字符串以将参数传递给url，并将其写入URLconnection对象。下面的例子将清楚我刚才所说的内容：

Uri.Builder builder = new Uri.Builder().appendQueryParameter("param1", value1) .appendQueryParameter("param2", value2); final String query = builder.build().getEncodedQuery(); URL url; HttpURLConnection connection = null; try{ url = new URL("http://www.example.com/directory/"); connection = (HttpURLConnection) url.openConnection(); OutputStream os = connection.getOutputStream(); BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8")); writer.write(query); writer.flush(); writer.close(); os.close(); connection.connect(); }

在服务器端，您只需使用参数名称（param1）检索参数值（value1）。此外，您可以使用.appendQueryParameter()方法添加所需数量的参数。

如何在android json中创建参数url

1 个答案: