Question

我发现<代码>hg status的输出量太高,无法追踪的名录。附录我有一个空洞的存放处,由<代码>git 和hg管理。因此,将有两个名录:.git和.hg。

<代码>git 状况 :

# Untracked files:
#   (use "git add <file>..." to include in what will be committed)
#
#   .hg/

<代码>hg 状况 :

? .git/HEAD
? .git/config
? .git/description
? .git/hooks/applypatch-msg.sample
? .git/hooks/commit-msg.sample
? .git/hooks/post-commit.sample
? .git/hooks/post-receive.sample
? .git/hooks/post-update.sample
? .git/hooks/pre-applypatch.sample
? .git/hooks/pre-commit.sample
? .git/hooks/pre-rebase.sample
? .git/hooks/prepare-commit-msg.sample
? .git/hooks/update.sample
? .git/info/exclude

是否有办法将其产出削减到如下一行?

? .git/

Answer 1

页: 1 档案

 syntax:glob
.git
.gitattributes
.gitignore

完全无视<代码>.git和其他与git有关的档案。

如果目标仅限于名录,则你仍可使用<代码>hgignore。如果在名录上添加一个秘密档案(,:

 syntax:regexp
 ^.git/(?!.hidden).+$

由于Mecurial根本不跟踪目录,这将忽略.git内的所有档案。除隐蔽外,实际上只显示一种情况。

You would also have the option -q/--quiet (to hg status), which hides untracked (unknown and ignored) files unless explicitly requested with -u/--unknown or -i/--ignored.
But that means .git would not even show up in that case.

但是,如果目标是将一般的hg status产出限制在名录上,而不是放在未跟踪档案的内容上,那么我认为这是不可能的:

git track content of files and since no content is added, it only mentions the top directory has "having no content added"
mercurial tracks files (not directories), hence the comprehensive list (of files).

Answer 2

这为我工作:

hg status -q

备选案文<代码>-q/---quiet藏匿未经跟踪(无人知晓和忽视)档案,除非有<代码>-u--un known或-i/--ignored的明确要求。

Answer 3

我理解的问题是如何减少产出: 我有同样的需要,但不想 un或忽略档案。如果你使用*NIX,那么你就使用灰尘,你只是想把产出作为你的问题国家加以削减,例如删除:

hg status | grep -v "^?s.git"

或例如没有显示已删除的档案

hg status | grep -v "^R"

Windows Power 壳牌相同:

hg status | Select-String -Pattern ("^[^R]")

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