python简单的wsgi文件上传脚本 - 出了什么问题？答案

【问题标题】：python simple wsgi file upload script - What is wrong?python简单的wsgi文件上传脚本 - 出了什么问题？
【发布时间】：2013-01-10 18:25:57
【问题描述】：

import os, cgi

#self_hosting script
tags = """<form enctype="multipart/form-data" action="save_file.py" method="post">
<p>File: <input type="file" name="file"></p>
<p><input type="submit" value="Upload"></p>
</form>"""

def Request(environ, start_response):
    # use cgi module to read data
    form = cgi.FieldStorage(fp=environ['wsgi.input'], environ=environ, keep_blank_values=True)

    try:
        fileitem = form['file']
    except KeyError:
        fileitem = None

    if fileitem and fileitem.file:
        fn = os.path.basename(fileitem.filename)
        with open(fn, 'wb') as f:
            data = fileitem.file.read(1024)
            while data:
                f.write(data)
                data = fileitem.file.read(1024)

            message = 'The file "' + fn + '" was uploaded successfully'

    else :
        message = 'please upload a file.'


    start_response('200 OK', [('Content-type','text/html')])

    return [message + "<br / >" + tags]

上面是我的 python wsgi 脚本，它接收文件并将其写入磁盘。但是，在执行时（选择文件）：

内部服务器错误处理此请求时发生错误。请求处理程序失败

Traceback (most recent call last):
  File "C:\Python26\Http\Isapi.py", line 110, in Request
    return Handler(Name)
  File "C:\Python26\Http\Isapi.py", line 93, in 
    "/apps/py/" : lambda P: RunWSGIWrapper(P),
  File "C:\Python26\Http\Isapi.py", line 86, in RunWSGIWrapper
    return RunWSGI(ScriptHandlers[Path])
  File "C:\Python26\Http\WSGI.py", line 155, in RunWSGI
    Result = Application(Environ, StartResponse)
  File "\\?\C:\Python26\html\save_file.py", line 13, in Request
    form = cgi.FieldStorage(fp=environ['wsgi.input'], environ=environ, keep_blank_values=True)
  File "C:\Python26\Lib\cgi.py", line 496, in __init__
    self.read_multi(environ, keep_blank_values, strict_parsing)
  File "C:\Python26\Lib\cgi.py", line 620, in read_multi
    environ, keep_blank_values, strict_parsing)
  File "C:\Python26\Lib\cgi.py", line 498, in __init__
    self.read_single()
  File "C:\Python26\Lib\cgi.py", line 635, in read_single
    self.read_lines()
  File "C:\Python26\Lib\cgi.py", line 657, in read_lines
    self.read_lines_to_outerboundary()
  File "C:\Python26\Lib\cgi.py", line 685, in read_lines_to_outerboundary
    line = self.fp.readline(1<<16)
AttributeError: 'module' object has no attribute 'readline'

在 wsgi 和 cgi 模块上相当愚蠢，我现在不知道要进步。有什么线索吗？

【问题讨论】：

标签： python post cgi wsgi

【解决方案1】：

environ['wsgi.input'] 是一个类似流的对象。您需要先将其缓存到类似对象的文件中，例如：tempfile.TemporaryFile 或StringIO（python3 中的io.BytesIO）：

from tempfile import TemporaryFile
import os, cgi

def read(environ):
    length = int(environ.get('CONTENT_LENGTH', 0))
    stream = environ['wsgi.input']
    body = TemporaryFile(mode='w+b')
    while length > 0:
        part = stream.read(min(length, 1024*200)) # 200KB buffer size
        if not part: break
        body.write(part)
        length -= len(part)
    body.seek(0)
    environ['wsgi.input'] = body
    return body

def Request(environ, start_response):
    # use cgi module to read data
    body = read(environ)
    form = cgi.FieldStorage(fp=body, environ=environ, keep_blank_values=True)
    # rest of your code

出于安全原因考虑屏蔽您传递给FieldStorage 的environ 值

【讨论】：

【解决方案2】：

自答：抱歉，我没有说这是 PyISAPIe 特有的问题。类文件对象environ['wsgi.input'] 没有readline() 方法，就像不同wsgi 实现中的其他environ 变量一样。

（低效的解决方法）是将 environ['wsgi.input'] 中的所有内容保存到 tempfile 并将其传递给 FieldStorage。

所以：

import tempfile, cgi
def some_wsgi_app(environ, start_response):
    temp_file = tempfile.TemporaryFile()
    temp_file.write(environ['wsgi.input'].read()) # or use buffered read()
    temp_file.seek(0)
    form = cgi.FieldStorage(fp=temp_file, environ=environ, keep_blank_values=True)
    # do_something #
    temp_file.close() #close and destroy temp file

    # ... start_response, return ... #

但是，如果用户上传的数据太大，上述示例将无法正常运行。

【讨论】：