bash 递归 Grep 仅显示总匹配计数
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Recursive Grep to show only total matchin count
提问by Mena Ortega
I'm trying to do a recursive grep search such as:
我正在尝试进行递归 grep 搜索，例如：
grep -r -c "foo" /some/directory
which give me output auch as:
这给了我输出，例如：
/some/directory/somefile_2013-04-08.txt:0
/some/directory/somefile_2013-04-09.txt:1
/some/directory/somefile_2013-04-10.txt:4
...etc
however, I would like to just get the total number of matches in all files, like:
但是，我只想获取所有文件中的匹配总数，例如：
Total matches: 5
I've played around with some other examples such as in this thread, although I can't seem to do what should be so simple.
我玩过一些其他的例子，比如在这个线程中，虽然我似乎无法做应该这么简单的事情。
回答by Halim Qarroum
grep -ro "foo" /some/directory | wc -l | xargs echo "Total matches :"
The -ooption of grepprints allthe existing occurences of a string in a file.
该-o的选项grep打印所有文件中的字符串的存在出现次数。
回答by dkmike
Just pipe to grep.
只需管道到grep。
grep -r "foo" /some/directory | grep -c "foo"
回答by John Kugelman
matches=$(grep -rch "foo" /some/directory | paste -sd+ - | bc)
echo "Total matches: $matches"
The paste/bc piece sums up a list of numbers. See: Bash command to sum a column of numbers.
paste/bc 部分总结了一个数字列表。请参阅：对一列数字求和的 Bash 命令。
回答by btanaka
If you only need the total, you can skip getting the count for each file, and just get a total number of matched lines:
如果您只需要总数，则可以跳过获取每个文件的计数，而只获取匹配行的总数：
grep -r "foo" /some/directory | wc -l
