如何将参数传递给java / android中的urlconnection？
发布时间：2020-12-15 08:32:39 所属栏目：Java 来源：网络整理
导读：我可以使用HttpUrlConnection建立连接.我的代码如下. client = new DefaultHttpClient();URL action_url = new URL(actionUrl);conn = (HttpURLConnection) action_url.openConnection();conn.setDoOutput(true);conn.setDoInput(true);conn.setRequestPrope
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我可以使用HttpUrlConnection建立连接.我的代码如下.
client = new DefaultHttpClient();
URL action_url = new URL(actionUrl);
conn = (HttpURLConnection) action_url.openConnection();
conn.setDoOutput(true);
conn.setDoInput(true);
conn.setRequestProperty("userType","2");
conn.setRequestProperty("Content-Type","application/x-www-form-urlencoded");
conn.setRequestMethod(HttpPost.METHOD_NAME);
DataOutputStream ds = new DataOutputStream(conn.getOutputStream());
String content = "username=username1&password=password11";
Log.v(TAG,"content: " + content);
ds.writeBytes(content);
ds.flush();
ds.close();
InputStream in = conn.getInputStream();//**getting filenotfound exception here.**
BufferedReader reader = new BufferedReader(
new InputStreamReader(in));
StringBuilder str1 = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
str1.append(line);
Log.v(TAG,"line:" + line);
}
in.close();
s = str1.toString();
获取filenotfound异常.不知道为什么？ 否则给我一些建议,通过代码将用户名和密码链接参数传递给网址. 解决方法
HTTPClient提供了一种更简单的方法来访问http资源,当你想要做的就是获取repsonse主体：
HttpGet httpGet = new HttpGet("http://domain.com/path?var1=bla&var2=foo");
HTTPResponse reponse = httpClient.execute(httpGet);
String responseBody = EntityUtils.toString(response.getEntity());
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