bookmark: complexity pull-push test to have deeper tree
This changeset only touch test.
The previous test was correct, it tested that the successors of an old bookmark
position was seen as a valid destination for bookmark.
However, a newer version is made for two reason:
(1) The new test check further. It check that the descendant of the successors
is a valid destination
(2) An ever more complexe test is needed to validate a future fix to issue 3680
Splitting complexification of the test and actual bugfix help to reduce the
noise in the bugfix changeset. Issue 3680 is NOT fixed by this changeset.
# Read the output of a "svn log --xml" command on stdin, parse it and
# print a subset of attributes common to all svn versions tested by
# hg.
import xml.dom.minidom, sys
def xmltext(e):
return ''.join(c.data for c
in e.childNodes
if c.nodeType == c.TEXT_NODE)
def parseentry(entry):
e = {}
e['revision'] = entry.getAttribute('revision')
e['author'] = xmltext(entry.getElementsByTagName('author')[0])
e['msg'] = xmltext(entry.getElementsByTagName('msg')[0])
e['paths'] = []
paths = entry.getElementsByTagName('paths')
if paths:
paths = paths[0]
for p in paths.getElementsByTagName('path'):
action = p.getAttribute('action')
path = xmltext(p)
frompath = p.getAttribute('copyfrom-path')
fromrev = p.getAttribute('copyfrom-rev')
e['paths'].append((path, action, frompath, fromrev))
return e
def parselog(data):
entries = []
doc = xml.dom.minidom.parseString(data)
for e in doc.getElementsByTagName('logentry'):
entries.append(parseentry(e))
return entries
def printentries(entries):
fp = sys.stdout
for e in entries:
for k in ('revision', 'author', 'msg'):
fp.write(('%s: %s\n' % (k, e[k])).encode('utf-8'))
for path, action, fpath, frev in sorted(e['paths']):
frominfo = ''
if frev:
frominfo = ' (from %s@%s)' % (fpath, frev)
p = ' %s %s%s\n' % (action, path, frominfo)
fp.write(p.encode('utf-8'))
if __name__ == '__main__':
data = sys.stdin.read()
entries = parselog(data)
printentries(entries)