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我遇到了这个问题，这个基本的，不成熟的我用Java编写的Web服务器。出于某种原因，不是仅向浏览器发送“200 OK”或“404 Not Found”，而是将它们写入正在检索的任何文件中。例如，如果我试图让index.html文件，我回到：错误返回文件w/Java Web服务器
....而不是浏览器试图编译HTML。 尝试获取图像时，情况会更糟糕，因为该服务器尝试向其添加“200 OK”&内容类型，该文件已损坏。任何人都可以提供任何有关如何发送状态行& content-type与实际的HTML/JPG文件分开的任何见解？当我注释掉“os.writeBytes（statusLine）”等等时，错误完全消失，但我仍然想将这些消息发送给浏览器...只是不合并文件。任何帮助将不胜感激。
import java.io.*;
import java.net.*;
import java.util.*;
final class HttpRequest implements Runnable {
final static String CRLF = "\r\n";
Socket socket;
// Constructor
public HttpRequest(Socket socket) throws Exception {
this.socket = socket;
}
// Implement the run() method of the Runnable interface.
public void run() {
try {
processRequest();
} catch (Exception e) {
System.out.println(e);
}
}
private static void sendBytes(FileInputStream fis, OutputStream os)
throws Exception {
// Construct a 1K buffer to hold bytes on their way to the socket.
byte[] buffer = new byte[1024];
int bytes = 0;
// Copy requested file into the socket's output stream.
while((bytes = fis.read(buffer)) != -1) {
os.write(buffer, 0, bytes);
}
}
private static String contentType(String fileName) {
if(fileName.endsWith(".htm") || fileName.endsWith(".html")) {
return "text/html";
}
if(fileName.endsWith(".jpeg") || fileName.endsWith(".jpg")) {
return "image/jpeg";
}
if(fileName.endsWith(".gif")) {
return "image/gif";
}
return "application/octet-stream";
}
private void processRequest() throws Exception {
// Get a reference to the socket's input and output streams.
InputStream is = socket.getInputStream();
DataOutputStream os = new DataOutputStream(socket.getOutputStream());
// Set up input stream filters.
BufferedReader br = new BufferedReader(new InputStreamReader(is));
// Get the request line of the HTTP request message.
String requestLine = new String(br.readLine());
// Display the request line.
System.out.println();
System.out.println(requestLine);
// Get and display the header lines.
String headerLine = null;
while ((headerLine = br.readLine()).length() != 0) {
System.out.println(headerLine);
}
// Extract the filename from the request line.
StringTokenizer tokens = new StringTokenizer(requestLine);
tokens.nextToken(); // skip over the method, which should be "GET"
String fileName = tokens.nextToken();
// Prepend a "." so that file request is within the current directory.
fileName = "." + fileName;
// Open the requested file.
FileInputStream fis = null;
boolean fileExists = true;
try {
fis = new FileInputStream(fileName);
} catch (FileNotFoundException e) {
fileExists = false;
}
// Construct the response message.
String statusLine = null;
String contentTypeLine = null;
String entityBody = null;
if (fileExists) {
statusLine = "200 OK" + CRLF;
contentTypeLine = "Content-type: " +
contentType(fileName) + CRLF;
} else {
statusLine = "404 NOT FOUND" + CRLF;
contentTypeLine = "Content Not Found!" + CRLF;
entityBody = "<HTML>" +
"<HEAD><TITLE>Not Found</TITLE></HEAD>" +
"<BODY>Not Found</BODY></HTML>";
}
// Send the status line.
os.writeBytes(statusLine);
// Send the content type line.
os.writeBytes(contentTypeLine);
// Send a blank line to indicate the end of the header lines.
os.writeBytes(CRLF);
// Send the entity body.
if (fileExists) {
sendBytes(fis, os);
fis.close();
} else {
os.writeBytes("File DNE: Content Not Found!");
}
// Close streams and socket.
os.close();
br.close();
socket.close();
}
public static void main(String[] args) throws Exception {
final ServerSocket ss = new ServerSocket(6789);
while (true)
new HttpRequest(ss.accept()).run();
}
}
我不太看到我可能错过了CRLF ...我有一个在状态线本身，以及在发送实体主体之前。你的意思是？ – 2012-02-21 04:24:42
请看我收录的链接。简而言之，您应该拥有HTTP/1.0 200 OK内容类型：文本/ html 嗨！虽然有比规范更多的东西。 –
MJB
2012-02-21 04:25:30