Check if a symlink is broken or circular

[davisn90210 at gmail.com]

Giampaolo Rodola' wrote:
> On 1 Dic, 00:10, "Martin v. L�wis" <mar... at v.loewis.de> wrote:
> > > I would like to know if such function would be correct for verifying
> > > if a link is broken and/or circular.
> >
> > > def isvalidlink(path):
> > >     assert os.path.islink(path)
> > >     try:
> > >         os.stat(path)
> > >     except os.error:
> > >         return 1
> > >     return 0
> >
> > You meant to flip the result values, right? 1 should mean that the
> > link is value, and 0 that it is not.
> >
> > Mostly. If the link is correct, but you don't have permission to stat
> > the target file, you get 0. OTOH, in that case, you have no way of
> > finding out whether the link *is* correct.
> >
> > Still, you could try to detect the errnos that indicate a problem
> > with the link itself, and pass all other errors through.
> >
> > Regards,
> > Martin
>
> Mmmm... do you mean something like this?
> Could it be ok?
>
>

I think he meant something like this (see below):

>
> import os, errno
>
> def isvalidlink(path):
>     assert os.path.lexists(path)
>     try:
>         os.stat(path)
>     except os.error, err:
>         # broken link
>         # "No such file or directory"
>         if err.errno == errno.ENOENT:
>             return 1

Right, except that it should be return 0 (or preferably return False)
unless you want inverted logic.

>         # circular link
>         # "Too many levels of symlinks"
>         elif err.errno == errno.ELOOP:
>             return 2

Change to return 1/True (inverted) or return 0/False (not inverted)

>         # something else occurred,
>         # assume it as invalid anyway
>         else:
>             return 3

Change "return 3" to "raise", so the exception gets propagated.

>     return 0

Again, if you do not want the logic to be inverted, use return 1/True
instead

--Nathan Davis