Python CGI脚本：OSError：[Errno 8] Exec格式错误

Question

When I send http://localhost:8888/cgi-bin/peoplecgi.py?action=Fetch&key=sue (sue is a valid key in the shelve) to the cgi script below I get the following (also I'm on OSX with Python 3.3). 当我发送http://localhost:8888/cgi-bin/peoplecgi.py?action=Fetch&key=sue （sue是搁置中的有效密钥）到下面的cgi脚本时，我得到以下内容（我也在OSX上）用Python 3.3）。 Any ideas what is going wrong? 出了什么问题？

127.0.0.1 - - [04/Feb/2014 10:38:41] "GET /cgi-bin/peoplecgi.py?action=Fetch&key=sue HTTP/1.1" 200 -
Traceback (most recent call last):
  File "/Library/Frameworks/Python.framework/Versions/3.3/lib/python3.3/http/server.py", line 1131, in run_cgi
    os.execve(scriptfile, args, env)
OSError: [Errno 8] Exec format error: '/Users/rich/Google Drive/Code/Python/PP4E/Preview/cgi-bin/peoplecgi.py'
127.0.0.1 - - [04/Feb/2014 10:38:41] CGI script exit status 0x7f00

I am working out of O'Reilly's Programming Python 4th Edition. 我正在研究O'Reilly的Programming Python第4版。 This is question is based off example 1-33. 这个问题是基于例子1-33。

weberver.py: weberver.py：

import os, sys
from http.server import HTTPServer, CGIHTTPRequestHandler

webdir = '/Users/rich/Google Drive/Code/Python/PP4E/Preview/'
port = 8888

os.chdir(webdir)
srvraddr = ("", port)
srvrobj = HTTPServer(srvraddr, CGIHTTPRequestHandler)
srvrobj.serve_forever()

peoplecgi.py peoplecgi.py

import cgi, shelve, sys, os
shelvename = 'class-shelve'
fieldnames = ('name', 'age', 'job', 'pay')

form = cgi.FieldStorage()
print('Content-type: text/html')
sys.path.insert(0, os.getcwd())

replyhtml="""
<html>
<title>People Input Form</title>
<body>
<form method=POST action="peoplecgi.py">
    <table>
    <tr><th>key<td><input type=text name=key value="%(key)s">
    $ROWS$
    </table>
    <p>
    <input type=submit value="Fetch", name=action>
    <input type=submit value="Update", name=action>
</form>
</body></html>
"""

rowhtml = '<tr><th>%s<td><input type=text name=%s value="%%(%s)s">\n'
rowshtml = ''
for fieldname in fieldnames:
    rowshtml += (rowhtml % ((fieldname,) * 3))

replyhtml = replyhtml.replace('$ROWS$', rowshtml)

def htmlize(adict):
    new = adict.copy()
    for field in fieldnames:
        value = new[field]
        new[field] = cgi.escape(repr(value))
    return new

def fetchRecord(db, form):
    try:
        key = form['key'].value
        record = db[key]
        fields = record.__dict__
        fields['key'] = key
    except:
        fields = dict.fromkeys(fieldnames, '?')
        fields['key'] = 'Missing or invalid key!'
    return fields

def updateRecord(db, form):
    if not 'key' in form:
        fields = dict.fromkeys(fieldnames, '?')
        fields['key'] = 'Missing key input!'
    else:
        key = form['key'].value
        if key in db:
            record = db[key]
        else:
            from person import Person
            record = Persion(name='?', age='?')
        for field in fieldnames:
            setattr(record, field, eval(form[field].value))
        db[key] = record
        fields = record.__dict__
        fields['key'] = key
    return fields

db = shelve.open(shelvename)
action = form['action'].value if 'action' in form else None
if action == 'Fetch':
    fields = updateRecord(db, form)
else:
    fields = dict.fromkeys(fieldnames, '?')
    fields['key'] = 'Missing or invalid action!'
db.close()
print(replyhtml % htmlize(fields))

Answer 1

The solution in this case was that I was using the wrong version of Python to run the script. 在这种情况下的解决方案是我使用错误版本的Python来运行脚本。 The default in osx 10.9 is Python 2.7.5. osx 10.9中的默认值是Python 2.7.5。 I wanted to run it with python 3.3, therefore my solution was simply to execute it with the appropriate version: #!/usr/local/bin/python3 我想用python 3.3运行它，因此我的解决方案只是用适当的版本执行它： #!/usr/local/bin/python3

Python CGI脚本：OSError：[Errno 8] Exec格式错误

问题描述

1 个解决方案

解决方案1
4 已采纳 2014-03-12 20:31:52

Python CGI脚本：OSError：[Errno 8] Exec格式错误

问题描述

1 个解决方案

解决方案1 4 已采纳 2014-03-12 20:31:52

解决方案1
4 已采纳 2014-03-12 20:31:52