sed's/pattern/subtitution/2g'真的有用嗎？

Question

爲什麼echo 'word word word word word word' | sed 's/[a-zA-Z]* /DELETED /2g'

打印word DELETED DELETED DELETED DELETED word？

爲什麼sed沒有替代最後的word？

Answer 1

因爲你的正則表達式有尾隨空格：

/[a-zA-Z]*/
     ^

它明確地尋找零個或多個字母，後面加一個空格。