我到处搜索,但找不到答案,有没有办法发出简单的 HTTP 请求?我想在我的一个网站上请求一个 PHP 页面/脚本,但我不想显示该网页.
I have searched everywhere but I couldn't find my answer, is there a way to make a simple HTTP request? I want to request a PHP page / script on one of my websites but I don't want to show the webpage.
如果可能的话,我什至想在后台进行(在 BroadcastReceiver 中)
If possible I even want to do it in the background (in a BroadcastReceiver)
推荐答案UPDATE
这是一个非常古老的答案.我绝对不会再推荐 Apache 的客户端了.而是使用:
UPDATE
This is a very old answer. I definitely won't recommend Apache's client anymore. Instead use either:
- 改造
- OkHttp
- 排球
- HttpUrlConnection
首先,请求访问网络的权限,在您的清单中添加以下内容:
First of all, request a permission to access network, add following to your manifest:
<uses-permission android:name="android.permission.INTERNET" />那么最简单的方法就是使用Android自带的Apache http客户端:
Then the easiest way is to use Apache http client bundled with Android:
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()); }如果您希望它在单独的线程上运行,我建议扩展 AsyncTask:
If you want it to run on separate thread I'd recommend extending 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.. } }然后您可以通过以下方式提出请求:
You then can make a request by:
new RequestTask().execute("stackoverflow");更多推荐
使用 android 发出 HTTP 请求
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