我到处都找过了,但我找不到我的答案,有没有一种方法可以做一个简单的HTTP请求?我想在我的一个网站上请求一个PHP页面/脚本,但我不想显示网页。
如果可能的话,我甚至想在后台(在一个BroadcastReceiver)做它
我到处都找过了,但我找不到我的答案,有没有一种方法可以做一个简单的HTTP请求?我想在我的一个网站上请求一个PHP页面/脚本,但我不想显示网页。
如果可能的话,我甚至想在后台(在一个BroadcastReceiver)做它
当前回答
更新
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。你可以用任意一种:
改造 OkHttp 截击 HttpUrlConnection
原来的答案
首先,申请访问网络的权限,在您的清单中添加以下内容:
<uses-permission android:name="android.permission.INTERNET" />
那么最简单的方法是使用Apache http客户端与Android捆绑:
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());
}
如果你想让它在单独的线程上运行,我建议扩展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..
}
}
然后,你可以通过以下方式提出请求:
new RequestTask().execute("http://stackoverflow.com");
其他回答
看看这个很棒的新库,它可以通过gradle获得:)
构建。Gradle:编译'com.apptakk.http_request:http-request:0.1.2'
用法:
new HttpRequestTask(
new HttpRequest("http://httpbin.org/post", HttpRequest.POST, "{ \"some\": \"data\" }"),
new HttpRequest.Handler() {
@Override
public void response(HttpResponse response) {
if (response.code == 200) {
Log.d(this.getClass().toString(), "Request successful!");
} else {
Log.e(this.getClass().toString(), "Request unsuccessful: " + response);
}
}
}).execute();
https://github.com/erf/http-request
除非你有明确的理由选择Apache HttpClient,否则你应该选择java.net.URLConnection。你可以在网上找到很多如何使用它的例子。
我们也改进了Android文档,因为你原来的帖子:http://developer.android.com/reference/java/net/HttpURLConnection.html
我们已经在官方博客http://android-developers.blogspot.com/2011/09/androids-http-clients.html上讨论了这些权衡
更新
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。你可以用任意一种:
改造 OkHttp 截击 HttpUrlConnection
原来的答案
首先,申请访问网络的权限,在您的清单中添加以下内容:
<uses-permission android:name="android.permission.INTERNET" />
那么最简单的方法是使用Apache http客户端与Android捆绑:
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());
}
如果你想让它在单独的线程上运行,我建议扩展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..
}
}
然后,你可以通过以下方式提出请求:
new RequestTask().execute("http://stackoverflow.com");
注意:与Android捆绑的Apache HTTP客户端现在已弃用,转而支持HttpURLConnection。请参阅Android开发者博客了解更多细节。
添加<uses-permission android:name="android.permission. "INTERNET" />到您的舱单。
然后你会像这样检索一个网页:
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();
}
我还建议在一个单独的线程上运行它:
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..
}
}
有关响应处理和POST请求的更多信息,请参阅文档。
private String getToServer(String service) throws IOException {
HttpGet httpget = new HttpGet(service);
ResponseHandler<String> responseHandler = new BasicResponseHandler();
return new DefaultHttpClient().execute(httpget, responseHandler);
}
问候