Sed替换整条生产线

Question

I posted this question on superuser a bit ago, but I haven't gotten an answer I like. Here it is, slightly modified.

I was hoping for a way to make sed replace the entire line with the replacement (rather than just the match) so I could do something like this:

sed -e "/$some_complex_regex_with_a_backref/\1/"

and have it only print the back-reference.

From this question , it seems like the way to do it is mess around with the regex to match the entire line, or use some other tool (like perl). Simply changing the regex to .*regex.* doesn't always work (as mentioned in that question). For example:

$ echo $regex
\([:alpha:]*\)day

$ echo "$phrase"
it is Saturday tomorrow
Warm outside...

$ echo "$phrase" | sed "s/$regex/\1/"
it is Satur tomorrow
Warm outside...

$ echo "$phrase" | sed "s/.*$regex.*/\1/"

Warm outside...

$ # what I'd like to have happen
$ echo "$phrase" | [[[some command or string of commands]]]
Satur
Warm outside...

I'm looking for the most concise way to replace the entire line (not just the matching part) assuming the following:

• The regex is in a variable, so can't be changed on a case by case basis.
• I'd like to do this without using perl or other beefier languages (sed, awk, grep, etc. are ok)
• The solution can't remove lines that don't match the original regex (sorry grep -o doesn't cut it).

What I thought of was the following (but it's ugly, so I'd like to know if there is something better):

$ sentinel=XXX
$ echo "$phrase" | sed "s/$regex/$sentinel\1$sentinel/" |
> sed "s/^.*$sentinel\(.*\)$sentinel.*$/\1/"

Answer 1

这可能对你有用：

sed '/'"$regex"'/!b;s//\n\1\n/;s/.*\n\(.*\)\n.*/\1/' file