regex - regex - match a word and replace content before it

Question

I have several files with this line:

<2-10 digits> ; word

I want to replace all of the digits that come before that word with something else. How can I do that?

Answer 1

 sed -i -e 's/.*word/something;word/g' <filename>

循环遍历目录中的多个文件。我假设 .txt 文件作为文件扩展名：

for i in `\ls -1 *.txt`
  do
    sed -i -e 's/.*word/something;word/g' $i
done

注意：sed -i将以交互方式修改文件。因此，在没有-i选项的情况下测试命令以检查这是您想要的，然后继续执行...

Answer 2

更新： sed 示例：

s="999 abc 1234 ; word 567"
echo $s | sed 's/^\(.* \)[0-9][0-9]*\( ; word.*\)$/\1something\2/g'

输出：

999 abc something ; word 567