我需要使用正则表达式替换多个文件中的内容。我是这样做的:
#!/usr/bin/env bash
find /path/to/dir \
-type f \
-name '*.txt' \
-exec perl -e 's/replaceWhat/replaceWith/ig' -pi '{}' \; \
-print \
| awk '{count = count+1 }; END { print count " file(s) handled" }'
echo "Done!"
此代码向用户显示其处理的文件数。但我怎么能算不是文件,而是替换?处理的每个文件都可以产生零,一个或多个替换正则表达式。
答案 0 :(得分:1)
您可以向-exec
添加额外的grep
来电,然后将匹配数传递给awk
,而不只是文件名:
#!/usr/bin/env bash
find /path/to/dir \
-type f \
-name '*.txt' \
-exec grep -c 'replaceWhat' '{}' \; \
-exec perl -e 's/replaceWhat/replaceWith/ig' -pi '{}' \; \
| awk '{count += $0 }; END { print count " replacement(s) made" }'
echo "Done!"
示例(替换"之前"与"之后")
$ tail -n +1 *.txt
==> 1.txt <==
before
foo
bar
==> 2.txt <==
foo
bar
baz
==> 3.txt <==
before
foo
before
bar
before
$ ./count_replacements.sh
4 replacement(s) handled
$ tail -n +1 *.txt
==> 1.txt <==
after
foo
bar
==> 2.txt <==
foo
bar
baz
==> 3.txt <==
after
foo
after
bar
after