极客教程Python 如何使用Python CGI编程进行文件上传？
要上传文件，HTML表单必须将enctype属性设置为multipart/form-data。文件类型的输入标记会创建一个“浏览”按钮。
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示例
<html>
<body>
<form enctype = "multipart/form-data"
action = "save_file.py" method = "post">
<p>File: <input type = "file" name = "filename" /></p>
<p><input type = "submit" value = "Upload" /></p>
</form>
</body>
</html>
输出
此代码的结果是以下表单-
File: Choose file
Upload
这是处理文件上传的save_file.py脚本-
# !/usr/bin/python
import cgi, os
import cgitb; cgitb.enable()
form = cgi.FieldStorage()
# Get filename here.
fileitem = form['filename']
# Test if the file was uploaded
if fileitem.filename:
# strip leading path from file name to avoid
# directory traversal attacks
fn = os.path.basename(fileitem.filename)
open('/tmp/' + fn, 'wb').write(fileitem.file.read())
message = '文件“' + fn + '”已成功上传'
else:
message = '未上传任何文件'
print """\
Content-Type: text/html\n
<html>
<body>
<p>%s</p>
</body>
</html>
""" % (message,)
如果在Unix/Linux上运行上面的脚本，则需要注意将文件分隔符替换为以下内容，否则在Windows机器上上述open（）语句应正常工作。
fn = os.path.basename(fileitem.filename.replace("\", "/" ))