view tests/test-glog-topological.t @ 35854:568917059243

testrunner: make reading of test times work with #testcases Due to a bug that will be fixed in the next patch, we never actually read back .testcases, so we didn't notice that it could not be parsed successfully when there are #testcases tests. The parsing failed on lines like "test-amend-subrepo.t (case obsstore-off) 32.420" because we used a simple string.split() call and expected all parts but the first to be floating point numbers (and "(case" isn't, for example). Fix by using a regex instead. Differential Revision: https://phab.mercurial-scm.org/D1960
author Martin von Zweigbergk <martinvonz@google.com>
date Wed, 31 Jan 2018 23:12:45 -0800
parents eb586ed5d8ce
children 89630d0b3e23
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This test file aims at test topological iteration and the various configuration it can has.

  $ cat >> $HGRCPATH << EOF
  > [ui]
  > logtemplate={rev}\n
  > EOF

On this simple example, all topological branch are displayed in turn until we
can finally display 0. this implies skipping from 8 to 3 and coming back to 7
later.

  $ hg init test01
  $ cd test01
  $ hg unbundle $TESTDIR/bundles/remote.hg
  adding changesets
  adding manifests
  adding file changes
  added 9 changesets with 7 changes to 4 files (+1 heads)
  new changesets bfaf4b5cbf01:916f1afdef90
  (run 'hg heads' to see heads, 'hg merge' to merge)

  $ hg log -G
  o  8
  |
  | o  7
  | |
  | o  6
  | |
  | o  5
  | |
  | o  4
  | |
  o |  3
  | |
  o |  2
  | |
  o |  1
  |/
  o  0
  

(display all nodes)

  $ hg log -G -r 'sort(all(), topo)'
  o  8
  |
  o  3
  |
  o  2
  |
  o  1
  |
  | o  7
  | |
  | o  6
  | |
  | o  5
  | |
  | o  4
  |/
  o  0
  

(display nodes filtered by log options)

  $ hg log -G -r 'sort(all(), topo)' -k '.3'
  o  8
  |
  o  3
  |
  ~
  o  7
  |
  o  6
  |
  ~

(revset skipping nodes)

  $ hg log -G --rev 'sort(not (2+6), topo)'
  o  8
  |
  o  3
  :
  o  1
  |
  | o  7
  | :
  | o  5
  | |
  | o  4
  |/
  o  0
  

(begin) from the other branch

  $ hg log -G -r 'sort(all(), topo, topo.firstbranch=5)'
  o  7
  |
  o  6
  |
  o  5
  |
  o  4
  |
  | o  8
  | |
  | o  3
  | |
  | o  2
  | |
  | o  1
  |/
  o  0