Create Specfile from a virtual file

[FrostyX]

Currently, it is possible to create an object of Specfile like this:

# using an absolute path
specfile = Specfile('/tmp/test.spec')

# using a relative path and a different sourcedir
specfile = Specfile('test.spec', sourcedir='/tmp/sources')

It would be nice to be able to create it also like this

with open('/tmp/test.spec', 'r') as fp:
    specfile = Specfile(fp)

This alone is not so interesting, but it allows for creating Specfile from a virtual file

from io import StringIO
fp = StringIO()
fp.write("Name: foo\nVersion: 1.0\nRelease: 1\n...")
specfile = Specfile(fp)

Of course, I am interested in the feature, not the exact syntax. My suggestion will probably have an issue with the .save() method so if it is done like

specfile = Specfile(fp=fp)

# or

specfile = Specfile.from_fp(fp)

It is all fine with me :-)

[nforro]

This shouldn't be a problem, I'd probably go with adding a fp argument to the constructor. However, sourcedir would have to be specified when using fp.

[majamassarini]

Still relevant but low priority.

[majamassarini]

@FrostyX is this still useful for you?

[FrostyX]

Still relevant but low priority.
@FrostyX is this still useful for you?

No problem with me. Somewhere (if I could only remember where) I wanted to do this and couldn't, so I workarounded it by saving the specfile from a variable to a temporary file and using Specfile the standard way.

I still think it is useful but I agree with the low priority since it can be easily workarounded.