Message77555
| Author |
francescor |
| Recipients |
francescor |
| Date |
2008年12月10日.16:27:45 |
| SpamBayes Score |
1.0866497e-06 |
| Marked as misclassified |
No |
| Message-id |
<1228926467.09.0.231688302794.issue4621@psf.upfronthosting.co.za> |
| In-reply-to |
| Content |
Each entry of a zip file, as read by the zipfile module, can be accessed
via a ZipInfo object. The filename attribute of ZipInfo is a string.
However, the read method of a ZipFile object expects a binary as
argument, or at least this is what I can deduct from the following behavior:
>>> import zipfile
>>> testzip = zipfile.ZipFile('test.zip')
>>> t1 = testzip.infolist()[0]
>>> t1.filename
'tést.xml'
>>> data = testzip.read(testzip.infolist()[0])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Python30\lib\zipfile.py", line 843, in read
return self.open(name, "r", pwd).read()
File "C:\Python30\lib\zipfile.py", line 883, in open
% (zinfo.orig_filename, fname))
zipfile.BadZipfile: File name in directory 'tést.xml' and header
b't\x82st.xml' differ.
The test.zip file is attached as help in reproducing this error. |
|
History
|
|---|
| Date |
User |
Action |
Args |
| 2008年12月10日 16:27:47 | francescor | set | recipients:
+ francescor |
| 2008年12月10日 16:27:47 | francescor | set | messageid: <1228926467.09.0.231688302794.issue4621@psf.upfronthosting.co.za> |
| 2008年12月10日 16:27:46 | francescor | link | issue4621 messages |
| 2008年12月10日 16:27:45 | francescor | create |
|