我在玩cmder (http://gooseberrycreative.com/cmder/)，我想计算html,css,java和javascript的行数。虽然上面的一些答案是有效的，或者grep中的模式没有-我在这里(https://unix.stackexchange.com/questions/37313/how-do-i-grep-for-multiple-patterns)发现我必须转义它

这就是我现在用的:

git ls-files | grep " \ (html css |。\ |。js |。java \)美元“| xargs厕所-

如果你想找到非空行总数，你可以使用AWK:

git ls-files | xargs cat | awk '/\S/{x++} END{print "非空行总数:"，x}'

它使用regex计算包含非空格字符的行数。

根据您是否希望包含二进制文件，有两种解决方案。

git grep --cached -al '' | xargs -P 4 cat | wc -l git grep --cached -Il '' | xargs -P 4 cat | wc -l "xargs -P 4" means it can read the files using four parallel processes. This can be really helpful if you are scanning very large repositories. Depending on capacity of the machine you may increase number of processes. -a, process binary files as text (Include Binary) -l '', show only filenames instead of matching lines (Scan only non empty files) -I, don't match patterns in binary files (Exclude Binary) --cached, search in index instead of in the work tree (Include uncommitted files)

从时钟1.68开始工作:

cloc——风险投资= git

: | git mktree | git diff --shortstat --stdin

Or:

git ls-tree @ | sed '1i\\' | git mktree --batch | xargs | git diff-tree --shortstat --stdin

Carl Norum的答案假设没有带空格的文件，IFS的一个字符是制表符和换行符。解决方案是用NULL字节终止该行。

 git ls-files -z | xargs -0 cat | wc -l