Java 为什么不成功;带套接字的HTTP GET请求;?
我想在我的Android应用程序上发送GET消息。之后我想收到回复为200 OK。但我没有做到。我收到了408个请求超时或什么都没有。你能帮我吗Java 为什么不成功;带套接字的HTTP GET请求;?,java,android,sockets,http,get,Java,Android,Sockets,Http,Get,我想在我的Android应用程序上发送GET消息。之后我想收到回复为200 OK。但我没有做到。我收到了408个请求超时或什么都没有。你能帮我吗 String requestmsg = "GET / HTTP/1.1\r\n"; requestmsg += "Host: www.ktu.edu.tr\r\n"; requestmsg += "Connection: keep-alive\r\n"; requestmsg += "Accept: text/html,
String requestmsg = "GET / HTTP/1.1\r\n";
requestmsg += "Host: www.ktu.edu.tr\r\n";
requestmsg += "Connection: keep-alive\r\n";
requestmsg += "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8\r\n";
requestmsg += "User-Agent: Mozilla/5.0 (Windows NT 6.3; WOW 64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/42.0.2311.135 Safari/537.36\r\n";
requestmsg += "Accept-Encoding: gzip, deflate, sdch\r\n";
requestmsg += "Accept-Language: en-US,en;q=0.8,en-GB;q=0.6\r\n";
DataOutputStream dos = null;
BufferedReader dis = null;
try {
Log.d("ClientActivity", "Connecting...");
String addr = InetAddress.getByName("www.ktu.edu.tr").getHostAddress().toString();
Socket socket = new Socket(addr, 80);
String data = "";
try {
Log.d("ClientActivity", "C: Sending command.");
dos = new DataOutputStream(socket.getOutputStream());
dis = new BufferedReader(new InputStreamReader(socket.getInputStream()));
dos.write(requestmsg.getBytes());
Log.i("ClientActivity", "RequestMsg Sent");
StringBuilder sb = new StringBuilder();
while ((data = dis.readLine()) != null) {
sb.append(data);
}
Log.i("ClientActivity", "C: Sent.");
Log.i("ClientActivity", "C: Received " + sb.toString());
} catch (Exception e) {
Log.e("ClientActivity", "S: Error", e);
}
socket.close();
Log.d("ClientActivity", "C: Closed.");
} catch (Exception e) {
Log.e("ClientActivity", "C: Error", e);
}
首先,您忘了在末尾添加另一行
\r\n
来正确完成请求
其次,您正在读取从服务器获得的所有信息,而不确定响应的实际长度
请使用HTTP库,或者通过学习学习正确使用HTTP协议。(套接字需要一些努力;您可能更喜欢不同的方法。标准的JSE方法是URL.openConnection。)
指定编码,否则它是默认的eNotPortable
new InputStreamReader(socket.getInputStream(), "Windows-1252"));
相反方向也一样:
requestmsg.getBytes("Windows-1252")
这是Windows Latin-1,即使指定了更有限的Latin-1“ISO-8859-1”,浏览器也会接受它。稍后检查编码
发送头最好不要说它准备使用压缩:
//requestmsg += "Accept-Encoding: gzip, deflate, sdch\r\n";
并用空行关闭请求标头:
requestmsg += "\r\n";
BufferedOutputStream比DataOutputStream IMHO更合适。事实上,正如其他人已经告诉您的那样,您需要一个结尾
requestmsg+=“\r\n”代码>
您最好删除行requestmsg+=“接受编码:gzip,deflate,sdch\r\n”
asreadLine()
无法处理zip内容
但是您的代码也会导致SocketTimeoutException
。你应该分开接
如果您访问我的服务器,并且与String host=“nl3.php.net”一起使用,那么您的代码就可以了代码>