bash - 带有空格的列表上的while循环

Question

我正在尝试运行一个 for 循环，它将替换作为列表传递的文件中的字符串。

例子：

Filename1

Filename2

File name3

File name4

我目前的代码是：

for i in `grep -Rl 'OLD' *`; do                                                                                                                 
> perl -pe 's/OLD/NEW/g' -pi "$i"
> done

Can't open File: No such file or directory.
Can't open name3: No such file or directory.
Can't open File: No such file or directory.
Can't open name3: No such file or directory.
Can't open File: No such file or directory.
Can't open name4: No such file or directory.

谢谢！

Answer 1

为了避免分词，你可以说：

grep -Rl 'OLD' * | while read i; do
  perl -pe 's/OLD/NEW/g' -pi "$i";
done

或者

while read i; do
  perl -pe 's/OLD/NEW/g' -pi "$i";
done < <(grep -Rl 'OLD' *)

Answer 2

尝试xargs使用sed

grep -Rl 'OLD' * -Z | xargs -0 sed -i 's/OLD/NEW/g'

请注意使用-Z选项 ofgrep以null终止输出，与xargs -0. 这将克服文件名中的空格问题。

Answer 3

您也可以使用 find 来做到这一点：

find -name 'File*' -exec sed -i 's,OLD,NEW,g' {} \;