Add 'list' command to repo.

This isn't a required command, but might be more discoverable for
repo newbies?

Change-Id: If357346f234774d42e04e024e65acdaf6dca6c62
This commit is contained in:
Doug Anderson 2011-03-14 16:07:42 -07:00
parent 37282b4b9c
commit fce89f218a

48
subcmds/list.py Normal file
View File

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#
# Copyright (C) 2011 The Android Open Source Project
#
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
#
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
from command import Command, MirrorSafeCommand
class List(Command, MirrorSafeCommand):
common = True
helpSummary = "List projects and their associated directories"
helpUsage = """
%prog [<project>...]
"""
helpDescription = """
List all projects; pass '.' to list the project for the cwd.
This is similar to running: repo forall -c 'echo "$REPO_PATH : $REPO_PROJECT"'.
"""
def Execute(self, opt, args):
"""List all projects and the associated directories.
This may be possible to do with 'repo forall', but repo newbies have
trouble figuring that out. The idea here is that it should be more
discoverable.
Args:
opt: The options. We don't take any.
args: Positional args. Can be a list of projects to list, or empty.
"""
projects = self.GetProjects(args)
lines = []
for project in projects:
lines.append("%s : %s" % (project.relpath, project.name))
lines.sort()
print '\n'.join(lines)