Question

我必须进行文件比较，但我希望您排除注释现在它给了我这样的结果。

目前我正在使用： diff -b -B [patch] [patch] 这让我：

< #<Location /bancochile>
< #        WeblogicHost bgri.wls.ri
< #        WeblogicPort 20015
< #        SetHandler weblogic-handler
< #</Location>
< 
45c39,43

Answer 1

我建议用bash：

diff -b -B <(grep -v '^#' patch1) <(grep -v '^#' patch2)

Answer 2

根据建议https://book.cakephp.org/3.0/en/development/routing.html#prefix-routing，GNU diff明确支持此目的：

diff -I '^[[:space:]]*[#]' -b -B old.cfg new.cfg

也就是说，如果你只是想知道是否有变化，那么使用cmp会更有效率：

if cmp -s <(grep -E -v '^[[:space:]]*[#]' <old.cfg) \
          <(grep -E -v '^[[:space:]]*[#]' <new.cfg); then
  echo "Excluding comments, these files are identical"
fi

我可以使用正则表达式忽略对diff的注释吗？

2 个答案: