使用 BaseHTTPRequestHandler 编写响应正文
我正在玩一点 Python 3.2.2，想编写一个简单的 Web 服务器来远程访问一些数据。该数据将由 Python 生成，因此我不想使用 SimpleHTTPRequestHandler，因为它是文件服务器，而是我自己的处理程序。
我从互联网上复制了一些示例，但我陷入困境，因为响应输出流拒绝写入正文内容。
这是我的代码：
import http.server
import socketserver
PORT = 8000
class MyHandler(http.server.BaseHTTPRequestHandler):
def do_HEAD(self):
self.send_response(200)
self.send_header("Content-type", "text/html")
self.end_headers()
def do_GET(self):
self.send_response(200)
self.send_header("Content-type", "text/html")
self.end_headers()
print(self.wfile)
self.wfile.write("<html><head><title>Title goes here.</title></head>")
self.wfile.write("<body><p>This is a test.</p>")
# If someone went to "http://something.somewhere.net/foo/bar/",
# then s.path equals "/foo/bar/".
self.wfile.write("<p>You accessed path: %s</p>" % self.path)
self.wfile.write("</body></html>")
self.wfile.close()
try:
server = http.server.HTTPServer(('localhost', PORT), MyHandler)
print('Started http server')
server.serve_forever()
except KeyboardInterrupt:
print('^C received, shutting down server')
server.socket.close()
编写响应正文的正确代码应该是什么？
多谢。
编辑：
错误是：
...
File "server.py", line 16, in do_GET
self.wfile.write("<html><head><title>Title goes here.</title></head>")
File "C:\Python32\lib\socket.py", line 297, in write
return self._sock.send(b)
TypeError: 'str' does not support the buffer interface
I'm playing a little with Python 3.2.2 and want to write a simple web server to access some data remotely. This data will be generated by Python so I don't want to use the SimpleHTTPRequestHandler as it's a file server, but a handler of my own.
I copied some example from the internet but I'm stuck because the response outputstream refuses to write the body content.
This is my code:
import http.server
import socketserver
PORT = 8000
class MyHandler(http.server.BaseHTTPRequestHandler):
def do_HEAD(self):
self.send_response(200)
self.send_header("Content-type", "text/html")
self.end_headers()
def do_GET(self):
self.send_response(200)
self.send_header("Content-type", "text/html")
self.end_headers()
print(self.wfile)
self.wfile.write("<html><head><title>Title goes here.</title></head>")
self.wfile.write("<body><p>This is a test.</p>")
# If someone went to "http://something.somewhere.net/foo/bar/",
# then s.path equals "/foo/bar/".
self.wfile.write("<p>You accessed path: %s</p>" % self.path)
self.wfile.write("</body></html>")
self.wfile.close()
try:
server = http.server.HTTPServer(('localhost', PORT), MyHandler)
print('Started http server')
server.serve_forever()
except KeyboardInterrupt:
print('^C received, shutting down server')
server.socket.close()
What should be a correct code for writing the response body?
Thanks a lot.
Edit:
The error is:
...
File "server.py", line 16, in do_GET
self.wfile.write("<html><head><title>Title goes here.</title></head>")
File "C:\Python32\lib\socket.py", line 297, in write
return self._sock.send(b)
TypeError: 'str' does not support the buffer interface
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评论（4）
Python3 中的字符串类型与 Python 2.x 中的字符串类型不同。使用
或将其转换为字节
In Python3 string is a different type than that in Python 2.x. Cast it into bytes using either
or
对于 Python 3，请在字符串文字前加上
b前缀：For Python 3, prefix the string literals with a
b:根据您的代码#comments，您可能正在寻找 self.headers.getheaders('referer')，即：
based on your code #comments you're probably looking for self.headers.getheaders('referer'), ie:
使用Python 3.X时只需使用这个
Just use this when using Python 3.X