google app引擎下载包含文件的文件

Question

好的，我有：

class Content(db.Model):
    code=db.TextProperty()

并且数据库中存储了3种不同的代码值。 如何创建一个zip文件，将三个代码值存储在3个单独的文件中，这些文件将被下载？

基于eric.f的答案：我重写了他的代码，使其可以执行我想要的操作：

    contents = db.GqlQuery("SELECT * FROM Content ORDER BY created DESC")
    output = StringIO.StringIO()
    with zipfile.ZipFile(output, 'w') as myzip:
        for content in contents:
            if content.code:
                code=content.code
            else:
                code=content.code2
            myzip.writestr('udacity_code'+`content.key().id()`, code)
    self.response.headers["Content-Type"] = "application/zip"
    self.response.headers['Content-Disposition'] = "attachment; filename=test.zip"
    self.response.out.write(output.getvalue())

我虽然出错了...

self.response.out.write(output.getvalue(), "utf-8")
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/StringIO.py", line 270, in getvalue
UnicodeDecodeError: 'ascii' codec can't decode byte 0xb4 in position 10: ordinal not in range(128)

Answer 1

import zipfile
import StringIO

output = StringIO.StringIO()

with zipfile.ZipFile(output, 'w') as myzip:
    myzip.writestr('file1.txt', 'aaaaaaaaa')
    myzip.writestr('file2.txt', 'bbbbbbbbb')
    myzip.writestr('file3.txt', 'ccccccccc')

然后做出响应，将output.getvalue()设置为内容，并设置标头，如下所示：

Content-type: application/zip
Content-disposition: attachment; filename=test.zip