NameError：全局名称未定义

Question

Hello My error is produced in generating a zip file. Can you inform what I should do?

main.py", line 2289, in get
    buf=zipf.read(2048)
NameError: global name 'zipf' is not defined

The complete code is as follows:

 def addFile(self,zipstream,url,fname):
     # get the contents          
     result = urlfetch.fetch(url)

     # store the contents in a stream
     f=StringIO.StringIO(result.content)
     length = result.headers['Content-Length']
     f.seek(0)

     # write the contents to the zip file
     while True:
       buff = f.read(int(length))
       if buff=="":break
       zipstream.writestr(fname,buff)
       return zipstream

 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-Length'] = len(output)    
    zipstream=StringIO.StringIO()
    file = zipfile.ZipFile(zipstream,"w")
    url = 'http://www.koolbusiness.com/list.kml'
    # repeat this for every URL that should be added to the zipfile
    file =self.addFile(file,url,"list.kml")
    # we have finished with the zip so package it up and write the directory
    file.close()
    zipstream.seek(0)
    # create and return the output stream
    self.response.headers['Content-Type'] ='application/zip'
    self.response.headers['Content-Disposition'] = 'attachment; filename="list.kmz"' 
    while True:
      buf=zipf.read(2048)
      if buf=="": break
    self.response.out.write(buf)

Answer 1

That is probably zipstream and not zipf . So replace that with zipstream and it might work.

Answer 2

i don't see where you declare zipf?

zipfile? Senthil Kumaran is probably right with zipstream since you seek(0) on zipstream before the while loop to read chunks of the mystery variable.

edit:

Almost certainly the variable is zipstream.

zipfile docs :

class zipfile.ZipFile(file[, mode[, compression[, allowZip64]]])

Open a ZIP file, where file can be either a path to a file (a string) or a file-like object. The mode parameter should be 'r' to read an existing file, 'w' to truncate and write a new file, or 'a' to append to an existing file. If mode is 'a' and file refers to an existing ZIP file, then additional files are added to it. If file does not refer to a ZIP file, then a new ZIP archive is appended to the file. This is meant for adding a ZIP archive to another file (such as python.exe).

your code:

zipsteam=StringIO.StringIO() 

create a file-like object using StringIO which is essentially a "memory file" read more in docs

file = zipfile.ZipFile(zipstream,w)

opens the zipfile with the zipstream file-like object in 'w' mode

url = 'http://www.koolbusiness.com/list.kml'
# repeat this for every URL that should be added to the zipfile
file =self.addFile(file,url,"list.kml")
# we have finished with the zip so package it up and write the directory
file.close()

uses the addFile method to retrieve and write the retrieved data to the file-like object and returns it. The variables are slightly confusing because you pass a zipfile to the addFile method which aliases as zipstream (confusing because we are using zipstream as a StringIO file-like object). Anyways, the zipfile is returned, and closed to make sure everything is "written".

It was written to our "memory file", which we now seek to index 0

zipstream.seek(0)

and after doing some header stuff, we finally reach the while loop that will read our "memory-file" in chunks

while True:
    buf=zipstream.read(2048)
    if buf=="": break
    self.response.out.write(buf)

Answer 3

You need to declare:

global zipf

right after your def get(self):

line. you are modifying a global variable, and this is the only way python knows what you are doing.