shell - sed:保留包含 block 模式的每个单词的最后一个 block

Question

愿意只保留最后一个包含点的单词 block (仍然保留不包含模式的单词):

sed 似乎只匹配最后一个单词:

$ echo "Here this.is.a.start is a very.nice.String  that.is.the.end yes " | sed 's/.*[ ]\([^ ]*\.\)\([^ ]*\)[ ].*/\2/g'
end

$ echo "Here this.is.a.start is a very.nice.String  that.is.the.end yes " | sed 's/.*[ ]\([^ ]*\.\)\([^ ]*\)[ ].*/\1/g'
that.is.the.

我应该怎样做才能得到这个结果？ :

echo "Here this.is.a.start is a very.nice.String  that.is.the.end yes " | sed s\\\g
Here start is a String  end yes

Answer 1

用\w 表示一个单词:

echo "Here this.is.a.start is a very.nice.String  that.is.the.end yes " |
 sed -E 's/\w*\.//g'