用android發出HTTP請求

我處處搜索但我找不到答案，有沒有辦法發出簡單的HTTP請求？ 我想在個人某個網站上請求PHP頁面/腳本，但我不想顯示該網頁。

若是可能的話我甚至想在後臺（在BroadcastReceiver中）這樣作

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#1樓

我使用Gson lib爲web服務建立了這個URL：

客戶：

public EstabelecimentoList getListaEstabelecimentoPorPromocao(){

        EstabelecimentoList estabelecimentoList  = new EstabelecimentoList();
        try{
            URL url = new URL("http://" +  Conexao.getSERVIDOR()+ "/cardapio.online/rest/recursos/busca_estabelecimento_promocao_android");
            HttpURLConnection con = (HttpURLConnection) url.openConnection();

            if (con.getResponseCode() != 200) {
                    throw new RuntimeException("HTTP error code : "+ con.getResponseCode());
            }

            BufferedReader br = new BufferedReader(new InputStreamReader((con.getInputStream())));
            estabelecimentoList = new Gson().fromJson(br, EstabelecimentoList.class);
            con.disconnect();

        } catch (IOException e) {
            e.printStackTrace();
        }
        return estabelecimentoList;
}

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#2樓

private String getToServer(String service) throws IOException {
    HttpGet httpget = new HttpGet(service);
    ResponseHandler<String> responseHandler = new BasicResponseHandler();
    return new DefaultHttpClient().execute(httpget, responseHandler);

}

問候

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#3樓

注意：如今不推薦使用與Android捆綁在一塊兒的Apache HTTP Client，而使用HttpURLConnection 。 有關詳細信息，請參閱Android開發人員博客 。

將<uses-permission android:name="android.permission.INTERNET" />到清單中。

而後，您將檢索以下所示的網頁：

URL url = new URL("http://www.android.com/");
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
try {
     InputStream in = new BufferedInputStream(urlConnection.getInputStream());
     readStream(in);
}
finally {
     urlConnection.disconnect();
}

我還建議在一個單獨的線程上運行它：

class RequestTask extends AsyncTask<String, String, String>{

@Override
protected String doInBackground(String... uri) {
    String responseString = null;
    try {
        URL url = new URL(myurl);
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        if(conn.getResponseCode() == HttpsURLConnection.HTTP_OK){
            // Do normal input or output stream reading
        }
        else {
            response = "FAILED"; // See documentation for more info on response handling
        }
    } catch (ClientProtocolException e) {
        //TODO Handle problems..
    } catch (IOException e) {
        //TODO Handle problems..
    }
    return responseString;
}

@Override
protected void onPostExecute(String result) {
    super.onPostExecute(result);
    //Do anything with response..
}
}

有關響應處理和POST請求的詳細信息，請參閱文檔 。

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#4樓

有一個線程：

private class LoadingThread extends Thread {
    Handler handler;

    LoadingThread(Handler h) {
        handler = h;
    }
    @Override
    public void run() {
        Message m = handler.obtainMessage();
        try {
            BufferedReader in = 
                new BufferedReader(new InputStreamReader(url.openStream()));
            String page = "";
            String inLine;

            while ((inLine = in.readLine()) != null) {
                page += inLine;
            }

            in.close();
            Bundle b = new Bundle();
            b.putString("result", page);
            m.setData(b);
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

        handler.sendMessage(m);
    }
}

---

#5樓

UPDATE

這是一個很是古老的答案。 我絕對不會再推薦Apache的客戶了。 而是使用：

• OkHttp
• HttpURLConnection的

原始答案

首先，請求訪問網絡的權限，在清單中添加如下內容：

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

那麼最簡單的方法是使用與Android捆綁的Apache http客戶端：

HttpClient httpclient = new DefaultHttpClient();
    HttpResponse response = httpclient.execute(new HttpGet(URL));
    StatusLine statusLine = response.getStatusLine();
    if(statusLine.getStatusCode() == HttpStatus.SC_OK){
        ByteArrayOutputStream out = new ByteArrayOutputStream();
        response.getEntity().writeTo(out);
        String responseString = out.toString();
        out.close();
        //..more logic
    } else{
        //Closes the connection.
        response.getEntity().getContent().close();
        throw new IOException(statusLine.getReasonPhrase());
    }

若是你想讓它在單獨的線程上運行，我建議擴展AsyncTask：

class RequestTask extends AsyncTask<String, String, String>{

    @Override
    protected String doInBackground(String... uri) {
        HttpClient httpclient = new DefaultHttpClient();
        HttpResponse response;
        String responseString = null;
        try {
            response = httpclient.execute(new HttpGet(uri[0]));
            StatusLine statusLine = response.getStatusLine();
            if(statusLine.getStatusCode() == HttpStatus.SC_OK){
                ByteArrayOutputStream out = new ByteArrayOutputStream();
                response.getEntity().writeTo(out);
                responseString = out.toString();
                out.close();
            } else{
                //Closes the connection.
                response.getEntity().getContent().close();
                throw new IOException(statusLine.getReasonPhrase());
            }
        } catch (ClientProtocolException e) {
            //TODO Handle problems..
        } catch (IOException e) {
            //TODO Handle problems..
        }
        return responseString;
    }

    @Override
    protected void onPostExecute(String result) {
        super.onPostExecute(result);
        //Do anything with response..
    }
}

而後您能夠經過如下方式提出請求：

new RequestTask().execute("http://stackoverflow.com");