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我的代码从 Google App Engine 生成一个zip 文件，用于在浏览器中加载，即 10 KB < 大小 < 1000 KB，我想知道我是否可以在这里使用 memcache，或者文件是否太大或已经缓存。在首先生成实际文件时，我已经使用了 memcache 并设置了缓存控制。

class KMZHandler(webapp.RequestHandler):
    def add_file(self, zip_file, url, file_name):
        """Fetch url, and add content as file_name to the zip file."""
        result = urlfetch.fetch(url)
        if not result.content:
            return
        zip_file.writestr(file_name, result.content)

    def get(self):
        """Attempt to create a zip file."""
        # you could set 'count' like this:
        # count = int(self.request.get('count', 1000))

        zipstream = StringIO.StringIO()
        zip_file = zipfile.ZipFile(zipstream, "w")

        # repeat this for every URL that should be added to the zipfile
        url = 'http://www.koolbusiness.com/list.kml'
        self.add_file(zip_file, url, "list.kml")

        # we have finished with the zip so package it up and write the directory
        zip_file.close()

        # set the headers...

        self.response.headers["Cache-Control"] = "public,max-age=%s" % 86400
        self.response.headers['Content-Type'] ='application/zip'
        self.response.headers['Content-Disposition'] = 'attachment;filename="list.kmz"'

        # create and return the output stream
        zipstream.seek(0)
        self.response.out.write(zipstream.read())
        zipstream.close()

这是使用 memcache 并创建实际文件的部分：

class KMLHandler(webapp.RequestHandler):
 def get(self):   
    self.response.headers["Cache-Control"] = "public,max-age=%s" % 86400
    start=datetime.datetime.now()-timedelta(days=20)
    count = int(self.request.get('count')) if not self.request.get('count')=='' else 1000        
    from google.appengine.api import memcache
    memcache_key = "ads"
    data = memcache.get(memcache_key)
    if data is None:
      a= Ad.all().filter("modified >", start).filter("url IN", ['www.koolbusiness.com']).filter("published =", True).order("-modified").fetch(count)
      memcache.set("ads", a)  
    else:
      a = data
    dispatch='templates/kml.html'
    template_values = {'a': a , 'request':self.request,}
    path = os.path.join(os.path.dirname(__file__), dispatch)
    output = template.render(path, template_values)
    self.response.headers['Content-Type'] = 'application/vnd.google-earth.kml+xml'
    self.response.headers['Content-Length'] = len(output)
    self.response.out.write(output)

感谢您的回答

4

1 回答 1

2

如果文件小于 1mb，并且每次请求都不会更改，那么对 KMZ 进行内存缓存可能会减少资源使用量。

KMZ 文件尚未在 memcache 中，但它可能在前端缓存中。当您生成 KML 时，您正在缓存来自查询的结果（请参阅 Nick 的博客以获取有关如何缓存实体的说明），但缓存不会考虑不同的值count或start。tf 那不重要，你也可以考虑直接memcaching KML（或KMZ）文件；如果您需要修复缓存策略很重要。

于 2011-04-17T14:08:30.570 回答