在GAE中从urlfetch打开一个文件 [英] Open a file from urlfetch in GAE

问题描述

我试图在GAE中打开一个文件，该文件使用 urlfetch（）获取。

到目前为止：

  from google.appengine.api import urlfetch 
 
 result = urlfetch。 fetch（'http://example.com/test.txt'）
 data = result.content 
 ## f = open（...）<  - 这是什么？ 
  

这看起来很奇怪，但BlobStore中有一个非常相似的可以将 data 写入BLOB文件：

  f = files.blobstore。使用files.open（f，'a'）作为数据创建（mime_type ='txt'，_blobinfo_uploaded_filename ='test'）
：
 data.write（result.content）
  

如何将 data 写入任意文件对象？

编辑：应该更清楚;我试图urlfetch 任何文件并在文件对象中打开 result.content 。所以它可能是一个.doc而不是.txt

解决方案

Yoy不必打开文件。您已收到data = result.content中的txt数据。

I'm trying to open a file in GAE that was retrieved using urlfetch().

Here's what I have so far:

from google.appengine.api import urlfetch

result = urlfetch.fetch('http://example.com/test.txt')
data = result.content
## f = open(...) <- what goes in here?

This might seem strange but there's a very similar function in the BlobStore that can write data to a blobfile:

f = files.blobstore.create(mime_type='txt', _blobinfo_uploaded_filename='test')
with files.open(f, 'a') as data:
    data.write(result.content)

How can I write data into an arbitrary file object?

Edit: Should've been more clear; I'm trying to urlfetch any file and open result.content in a file object. So it might be a .doc instead of a .txt

解决方案

Yoy do not have to open a file. You have received the txt data in data = result.content.

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