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java - 如何将参数传递给java/android中的urlconnection?

转载 作者:太空宇宙 更新时间:2023-11-03 12:25:39 25 4
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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参数传递给 url..

最佳答案

HTTPClient 提供了一种访问 http 资源的更简单的方法,当您只想获取响应正文时:

HttpGet httpGet = new HttpGet("http://domain.com/path?var1=bla&var2=foo");
HTTPResponse reponse = httpClient.execute(httpGet);
String responseBody = EntityUtils.toString(response.getEntity());

关于java - 如何将参数传递给java/android中的urlconnection?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/2437854/

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