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Use ',' instead of ':' or ';' to separate host patterns

The earlier-recommended "pat1:pat2:pat3[x:y]" notation doesn't work well
with IPv6 addresses, so we recommend ',' as a separator instead. We know
that commas can't occur within a pattern, so we can just split on it.

We still have to accept the "foo:bar" notation because it's so commonly
used, but we issue a deprecation warning for it.

Fixes #12296
Closes #12404
Closes #12329
This commit is contained in:
Abhijit Menon-Sen 2015-09-17 14:42:16 +05:30
parent 3db8070aa3
commit 2405861a9e

View file

@ -149,23 +149,6 @@ class Inventory(object):
results.append(item) results.append(item)
return results return results
def _split_pattern(self, pattern):
"""
takes e.g. "webservers[0:5]:dbservers:others"
and returns ["webservers[0:5]", "dbservers", "others"]
"""
term = re.compile(
r'''(?: # We want to match something comprising:
[^:\[\]] # (anything other than ':', '[', or ']'
| # ...or...
\[[^\]]*\] # a single complete bracketed expression)
)* # repeated as many times as possible
''', re.X
)
return [x for x in term.findall(pattern) if x]
def get_hosts(self, pattern="all", ignore_limits_and_restrictions=False): def get_hosts(self, pattern="all", ignore_limits_and_restrictions=False):
""" """
Takes a pattern or list of patterns and returns a list of matching Takes a pattern or list of patterns and returns a list of matching
@ -173,14 +156,6 @@ class Inventory(object):
or applied subsets or applied subsets
""" """
# Enumerate all hosts matching the given pattern (which may be
# either a list of patterns or a string like 'pat1:pat2').
if isinstance(pattern, list):
pattern = ':'.join(pattern)
if ';' in pattern or ',' in pattern:
display.deprecated("Use ':' instead of ',' or ';' to separate host patterns", version=2.0, removed=True)
patterns = self._split_pattern(pattern) patterns = self._split_pattern(pattern)
hosts = self._evaluate_patterns(patterns) hosts = self._evaluate_patterns(patterns)
@ -197,6 +172,59 @@ class Inventory(object):
return hosts return hosts
def _split_pattern(self, pattern):
"""
Takes a string containing host patterns separated by commas (or a list
thereof) and returns a list of single patterns (which may not contain
commas). Whitespace is ignored.
Also accepts ':' as a separator for backwards compatibility, but it is
not recommended due to the conflict with IPv6 addresses and host ranges.
Example: 'a,b[1], c[2:3] , d' -> ['a', 'b[1]', 'c[2:3]', 'd']
"""
if isinstance(pattern, list):
pattern = ','.join(pattern)
patterns = []
if ';' in pattern:
display.deprecated("Use ',' instead of ':' or ';' to separate host patterns", version=2.0, removed=True)
# If it's got commas in it, we'll treat it as a straightforward
# comma-separated list of patterns.
elif ',' in pattern:
patterns = re.split('\s*,\s*', pattern)
# If it doesn't, it could still be a single pattern. This accounts for
# non-separator uses of colons: IPv6 addresses and [x:y] host ranges.
else:
(base, port) = parse_address(pattern, allow_ranges=True)
if base:
patterns = [pattern]
# The only other case we accept is a ':'-separated list of patterns.
# This mishandles IPv6 addresses, and is retained only for backwards
# compatibility.
else:
patterns = re.findall(
r'''(?: # We want to match something comprising:
[^\s:\[\]] # (anything other than whitespace or ':[]'
| # ...or...
\[[^\]]*\] # a single complete bracketed expression)
)+ # occurring once or more
''', pattern, re.X
)
if len(patterns) > 1:
display.deprecated("Use ',' instead of ':' or ';' to separate host patterns", version=2.0)
return [p.strip() for p in patterns]
def _evaluate_patterns(self, patterns): def _evaluate_patterns(self, patterns):
""" """
Takes a list of patterns and returns a list of matching host names, Takes a list of patterns and returns a list of matching host names,
@ -249,7 +277,7 @@ class Inventory(object):
The pattern may be: The pattern may be:
1. A regex starting with ~, e.g. '~[abc]*' 1. A regex starting with ~, e.g. '~[abc]*'
2. A shell glob pattern with ?/*/[chars]/[!chars], e.g. 'foo' 2. A shell glob pattern with ?/*/[chars]/[!chars], e.g. 'foo*'
3. An ordinary word that matches itself only, e.g. 'foo' 3. An ordinary word that matches itself only, e.g. 'foo'
The pattern is matched using the following rules: The pattern is matched using the following rules: