Question

我目前正在编写一个python程序，使用sys.stdout.print()打印HTML页面，脚本名称为hellocgi.py，它位于Mac OS X桌面上的cgi-bin目录中。

以下是代码：

import sys
sys.stdout.write("Content-type: text/html \r\n\r\n")
sys.stdout.write("<!doctype html><html><head><title>Hello World</title></head>")
sys.stdout.write("<body><h2>Hello CGI</h2></body></html>")

然后我在cgi-bin目录中运行命令python -m CGIHTTPServer，当我要访问url localhost:8000/hellocgi.py时，我得到的是firefox中的源代码。

如果我尝试在cgi-bin之外运行python -m CGIHTTPServer，我会在网址localhost:8000/cgi-bin/上收到错误：

错误回复

错误代码403。

消息：CGI脚本不是普通文件（＆＃39; / cgi-bin /＆＃39;）。

错误代码说明：403 =请求被禁止 - 授权将   没有帮助。

在网址localhost:8000/cgi-bin/hellocgi.py上，firefox提示我下载源代码。

Python版本：2.7.5 Mac OS X：10.9

python -m CGIHTTPServer日志：

Serving HTTP on 0.0.0.0 port 8000 ...
127.0.0.1 - - [24/Jun/2014 18:34:49] "GET / HTTP/1.1" 200 -
127.0.0.1 - - [24/Jun/2014 18:34:51] code 403, message CGI script is not a plain file ('/cgi-bin/')
127.0.0.1 - - [24/Jun/2014 18:34:51] "GET /cgi-bin/ HTTP/1.1" 403 -
127.0.0.1 - - [24/Jun/2014 18:36:01] "GET /cgi-bin/hellocgi.py HTTP/1.1" 200 -
Traceback (most recent call last):
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/CGIHTTPServer.py", line 253, in run_cgi
    os.execve(scriptfile, args, env)
OSError: [Errno 13] Permission denied
127.0.0.1 - - [24/Jun/2014 18:36:01] CGI script exit status 0x7f00
^CTraceback (most recent call last):
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/runpy.py", line 162, in _run_module_as_main
    "__main__", fname, loader, pkg_name)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/runpy.py", line 72, in _run_code
    exec code in run_globals
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/CGIHTTPServer.py", line 378, in <module>
    test()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/CGIHTTPServer.py", line 374, in test
    SimpleHTTPServer.test(HandlerClass, ServerClass)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/SimpleHTTPServer.py", line 216, in test
    BaseHTTPServer.test(HandlerClass, ServerClass)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/BaseHTTPServer.py", line 599, in test
    httpd.serve_forever()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/SocketServer.py", line 236, in serve_forever
    poll_interval)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/SocketServer.py", line 155, in _eintr_retry
    return func(*args)

任何帮助？

Answer 1

#!/usr/bin/python
import sys
sys.stdout.write("Content-type: text/html \r\n\r\n")
sys.stdout.write("<!doctype html><html><head><title>Hello CGI</title></head>")
sys.stdout.write("<body><h2>Hello CGI</h2></body></html>")

并将preision添加到文件chmod + x querys.py

CGIHTTPServer和Python脚本

1 个答案: