Mercurial > hg
view mercurial/dicthelpers.py @ 22304:5678b0e3608f
templater: enable alias predicates to be used in "revset()" function
Before this patch, predicates defined in "[revsetalias]" can't be used
in the query specified to template function "revset()", because:
- "revset()" uses "localrepository.revs()" to get query result, but
- "localrepository.revs()" passes "None" as "ui" to "revset.match()", then
- "revset.match()" can't recognize any alias predicates
To enable alias predicates to be used in "revset()" function, this
patch invokes "revset.match()" directly with "repo.ui".
This patch doesn't make "localrepository.revs()" pass "self.ui" to
"revset.match()", because this may be intentional implementation to
prevent alias predicates from shadowing built-in ones and breaking
functions internally using "localrepository.revs()".
Even if it isn't intentional one, the check for shadowing should be
implemented (maybe on default branch) before fixing it for safety.
author | FUJIWARA Katsunori <foozy@lares.dti.ne.jp> |
---|---|
date | Sat, 23 Aug 2014 21:23:02 +0900 |
parents | ed46c2b98b0d |
children |
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# dicthelpers.py - helper routines for Python dicts # # Copyright 2013 Facebook # # This software may be used and distributed according to the terms of the # GNU General Public License version 2 or any later version. def diff(d1, d2, default=None): '''Return all key-value pairs that are different between d1 and d2. This includes keys that are present in one dict but not the other, and keys whose values are different. The return value is a dict with values being pairs of values from d1 and d2 respectively, and missing values treated as default, so if a value is missing from one dict and the same as default in the other, it will not be returned.''' res = {} if d1 is d2: # same dict, so diff is empty return res for k1, v1 in d1.iteritems(): v2 = d2.get(k1, default) if v1 != v2: res[k1] = (v1, v2) for k2 in d2: if k2 not in d1: v2 = d2[k2] if v2 != default: res[k2] = (default, v2) return res def join(d1, d2, default=None): '''Return all key-value pairs from both d1 and d2. This is akin to an outer join in relational algebra. The return value is a dict with values being pairs of values from d1 and d2 respectively, and missing values represented as default.''' res = {} for k1, v1 in d1.iteritems(): if k1 in d2: res[k1] = (v1, d2[k1]) else: res[k1] = (v1, default) if d1 is d2: return res for k2 in d2: if k2 not in d1: res[k2] = (default, d2[k2]) return res