[英]How to take a zipfile as a File object in django?
I have a folder named 'src' of which I create a zip file named zf. 我有一个名为“ src”的文件夹,其中创建了一个名为zf的zip文件。 I want to pass zf as a FileField object to newobj.fd.
我想将zf作为FileField对象传递给newobj.fd。
import zipfile
from django.core.files import File
f='s1.zip'
zf = zipfile.ZipFile(f, "w")
for dirname, subdirs, files in os.walk(src):
zf.write(dirname)
for filename in files:
zf.write(os.path.join(dirname, filename))
zf.open(os.path.join(dirname, filename))
newobj.fd= File(zf)
I do this thing for a text file and it works: 我对一个文本文件执行此操作,它可以正常工作:
f=file('text.txt')
newobj.fd2=File(f)
f.close()
How do the same thing for a zipfile? zipfile怎么做?
import zipfile
from django.core.files import File
f='s1.zip'
zf = zipfile.ZipFile(f, "w")
for dirname, subdirs, files in os.walk(src):
zf.write(dirname)
for filename in files:
zf.write(os.path.join(dirname, filename))
zf.close()
newobj.fd = File(f)
What does this give? 这有什么作用?
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