我怎样才能在python cgi中找到上传的文件名
我制作了如下简单的网络服务器。
import BaseHTTPServer, os, cgi
import cgitb; cgitb.enable()
html = """
<html>
<body>
<form action="" method="POST" enctype="multipart/form-data">
File upload: <input type="file" name="upfile">
<input type="submit" value="upload">
</form>
</body>
</html>
"""
class Handler(BaseHTTPServer.BaseHTTPRequestHandler):
def do_GET(self):
self.send_response(200)
self.send_header("content-type", "text/html;charset=utf-8")
self.end_headers()
self.wfile.write(html)
def do_POST(self):
ctype, pdict = cgi.parse_header(self.headers.getheader('content-type'))
if ctype == 'multipart/form-data':
query = cgi.parse_multipart(self.rfile, pdict)
upfilecontent = query.get('upfile')
if upfilecontent:
# i don't know how to get the file name.. so i named it 'tmp.dat'
fout = file(os.path.join('tmp', 'tmp.dat'), 'wb')
fout.write (upfilecontent[0])
fout.close()
self.do_GET()
if __name__ == '__main__':
server = BaseHTTPServer.HTTPServer(("127.0.0.1", 8080), Handler)
print('web server on 8080..')
server.serve_forever()
在BaseHTTPRequestHandler的do_Post方法中,我成功获取了上传的文件数据。
但我不知道如何获取上传文件的原始名称。 self.rfile.name 只是一个“套接字” 如何获取上传的文件名?
i made simple web server like below.
import BaseHTTPServer, os, cgi
import cgitb; cgitb.enable()
html = """
<html>
<body>
<form action="" method="POST" enctype="multipart/form-data">
File upload: <input type="file" name="upfile">
<input type="submit" value="upload">
</form>
</body>
</html>
"""
class Handler(BaseHTTPServer.BaseHTTPRequestHandler):
def do_GET(self):
self.send_response(200)
self.send_header("content-type", "text/html;charset=utf-8")
self.end_headers()
self.wfile.write(html)
def do_POST(self):
ctype, pdict = cgi.parse_header(self.headers.getheader('content-type'))
if ctype == 'multipart/form-data':
query = cgi.parse_multipart(self.rfile, pdict)
upfilecontent = query.get('upfile')
if upfilecontent:
# i don't know how to get the file name.. so i named it 'tmp.dat'
fout = file(os.path.join('tmp', 'tmp.dat'), 'wb')
fout.write (upfilecontent[0])
fout.close()
self.do_GET()
if __name__ == '__main__':
server = BaseHTTPServer.HTTPServer(("127.0.0.1", 8080), Handler)
print('web server on 8080..')
server.serve_forever()
In the do_Post method of BaseHTTPRequestHandler, i got the uploaded file data successfully.
But i can't figure out how to get the original name of the uploaded file.
self.rfile.name is just a 'socket'
How can i get the uploaded file name?
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您在那里作为起点使用的代码相当损坏(例如,查看
global rootnode
,其中名称rootnode
被使用无处 - 显然是一半- 编辑过的源代码,而且做得很糟糕)。无论如何,您在
POST
中使用“客户端”形式是什么?它如何设置upfile
字段?为什么不使用正常的 FieldStorage 方法,如 Python 文档?这样,您可以使用相应字段的
.file
属性来读取要读取的类似文件的对象,或其.value
属性来读取内存中的所有内容,并将其作为字符串获取,加上字段的.filename
属性即可知道上传的文件的名称。有关FieldStorage
的更详细但简洁的文档位于此处。编辑:现在OP已经编辑了Q来澄清,我看到了问题:
BaseHTTPServer
确实没有根据CGI规范设置环境,因此cgi
模块对此不太有用。不幸的是,环境设置的唯一简单方法是从 CGIHTTPServer.py 中窃取和破解一大段代码(无意重用,因此需要,叹息,复制和粘贴代码),例如...:这可以进一步大大简化,但必须在该任务上花费一些时间和精力:-(。
有了这个
populenv
函数,我们可以重新编码:...并实时从此幸福快乐;-)。 (当然,使用任何像样的 WSGI 服务器,甚至演示服务器,会容易得多,但是这个练习对 CGI 及其内部结构有指导意义;-)。
Pretty broken code you're using there as a starting point (e.g. look at that
global rootnode
where namerootnode
is used nowhere -- clearly half-edited source, and badly at that).Anyway, what form are you using "client-side" for the
POST
? How does it set thatupfile
field?Why aren't you using the normal
FieldStorage
approach, as documented in Python's docs? That way, you could use the.file
attribute of the appropriate field to get a file-like object to read, or its.value
attribute to read it all in memory and get it as a string, plus the.filename
attribute of the field to know the uploaded file's name. More detailed, though concise, docs onFieldStorage
, are here.Edit: now that the OP has edited the Q to clarify, I see the problem:
BaseHTTPServer
does not set the environment according to the CGI specs, so thecgi
module isn't very usable with it. Unfortunately the only simple approach to environment setting is to steal and hack a big piece of code fromCGIHTTPServer.py
(wasn't intented for reuse, whence the need for, sigh, copy and paste coding), e.g....:This could be substantially simplified further, but not without spending some time and energy on that task:-(.
With this
populenv
function at hand, we can recode:...and live happily ever after;-). (Of course, using any decent WSGI server, or even the demo one, would be much easier, but this exercise is instructive about CGI and its internals;-).
通过使用 cgi.FieldStorage 您可以轻松提取文件名。检查下面的示例:
By using cgi.FieldStorage you can easily extract the filename. Check the example below:
...或者使用您自己的 cgi.parse_multipart 版本,特别是修复此问题:
...or use your own version of cgi.parse_multipart, especially fixing this: