Francesco Ricciardi <francesco.riccia...@hp.com> added the comment:
If that is what is requested, then the manual entry for ZipFile.read
must be corrected, because it states:
"ZipFile.read(name[, pwd]) .... name is the name of the file in the
archive, or a ZipInfo object."
However, Eddie, you haven't tried what you suggested, because this is
what you would get:
>>> import zipfile
>>> testzip = zipfile.ZipFile('test.zip')
>>> t1 = testzip.infolist()[0]
>>> t1.filename
'tést.xml'
>>> data = testzip.read(t1.filename)
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.
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<http://bugs.python.org/issue4621>
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