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
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
其他回答
这是android中HTTP Get/POST请求的新代码。HTTPClient已被废弃,可能无法使用,因为在我的情况下。
首先在build.gradle中添加两个依赖:
compile 'org.apache.httpcomponents:httpcore:4.4.1'
compile 'org.apache.httpcomponents:httpclient:4.5'
然后在ASyncTask in doBackground方法中编写此代码。
URL url = new URL("http://localhost:8080/web/get?key=value");
HttpURLConnection urlConnection = (HttpURLConnection)url.openConnection();
urlConnection.setRequestMethod("GET");
int statusCode = urlConnection.getResponseCode();
if (statusCode == 200) {
InputStream it = new BufferedInputStream(urlConnection.getInputStream());
InputStreamReader read = new InputStreamReader(it);
BufferedReader buff = new BufferedReader(read);
StringBuilder dta = new StringBuilder();
String chunks ;
while((chunks = buff.readLine()) != null)
{
dta.append(chunks);
}
}
else
{
//Handle else
}
有一根线:
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);
}
}
最简单的方法是使用名为Volley的Android库
Volley offers the following benefits: Automatic scheduling of network requests. Multiple concurrent network connections. Transparent disk and memory response caching with standard HTTP cache coherence. Support for request prioritization. Cancellation request API. You can cancel a single request, or you can set blocks or scopes of requests to cancel. Ease of customization, for example, for retry and backoff. Strong ordering that makes it easy to correctly populate your UI with data fetched asynchronously from the network. Debugging and tracing tools.
你可以发送一个http/https请求,简单如下:
// Instantiate the RequestQueue.
RequestQueue queue = Volley.newRequestQueue(this);
String url ="http://www.yourapi.com";
JsonObjectRequest request = new JsonObjectRequest(url, null,
new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
if (null != response) {
try {
//handle your response
} catch (JSONException e) {
e.printStackTrace();
}
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
}
});
queue.add(request);
在这种情况下,你不必考虑“运行在后台”或“使用缓存”自己,因为所有这些都已经完成了Volley。
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
我做了这个webservice请求URL,使用一个Gson库:
客户:
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;
}