* Florian Weimer:
I'm going to add something like the following:
Testing reveals that this not what APT does.
I'm inclined to handle ~ versions only if python-apt is installed.
Would this be acceptable to you?
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On Sat, Jan 14, 2006 at 03:14:55PM +0100, Florian Weimer wrote:
I'm inclined to handle ~ versions only if python-apt is installed.
Would this be acceptable to you?
I think so. Please have your package suggest python-apt and emit a
meaningful warning if python-apt is not found and a ~ version is
* Marc Haber:
What happens if there are multiple ~?
They are processed in order.
dpkg --compare-versions handles ~ correctly.
And APT? Does it behave differently? (There are differences between
the two in the area of epoch handling.)
I'm going to add something like the following:
-
Package: debsecan
Version: 0.3.4
Severity: normal
Hi,
debsecan complains
invalid version 1.2.9-1~zg1 of package $PACKAGE
The version is, however, correct. This should be fixed.
Greetings
Marc
-- System Information:
Debian Release: testing/unstable
APT prefers unstable
APT policy: (500,
* Marc Haber:
debsecan complains
invalid version 1.2.9-1~zg1 of package $PACKAGE
The version is, however, correct. This should be fixed.
How? Is there an official description of the ~ semantics?
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On Fri, Jan 06, 2006 at 10:55:28AM +0100, Florian Weimer wrote:
* Marc Haber:
debsecan complains
invalid version 1.2.9-1~zg1 of package $PACKAGE
The version is, however, correct. This should be fixed.
How? Is there an official description of the ~ semantics?
I didn't find any
* Marc Haber:
On Fri, Jan 06, 2006 at 10:55:28AM +0100, Florian Weimer wrote:
* Marc Haber:
debsecan complains
invalid version 1.2.9-1~zg1 of package $PACKAGE
The version is, however, correct. This should be fixed.
How? Is there an official description of the ~ semantics?
I didn't
On Fri, Jan 06, 2006 at 12:33:09PM +0100, Florian Weimer wrote:
* Marc Haber:
I didn't find any official description short of #150739 and #93386.
The semantics are, that 1.0-1~1 is smaller than 1.0-1 but greater than
1.0-0.
And this does indeed result in a linear ordering?
Probably.
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