When we match a pattern using sed, the matched pattern is stored in the "ampersand" (&) variable. IS there a way to replace a character in this matched pattern using the ampersand itself ?
For example, if & contains the string "apple1", how can I use & to make the string to "apple2" (i.e replace 1 by 2) ?
If I guessed right, what you want to do is to apply a subsitution in a pattern matched. You can't do that using &. You want to do this instead:
echo apple1 apple3 apple1 apple2 botemo1 | sed '/apple./ { s/apple1/apple2/g; }'
This means that you want to execute the command substitution only on the lines that matches the pattern /apple./.
You can also use a capture group. A capture is used to grab a part of the match and save it into an auxiliary variable, that is named numerically in the order that the capture appears.
echo apple1 | sed -e 's/\(a\)\(p*\)\(le\)1/\1\2\32/g'
We used three captures:
- The first one, stored in
\1, contains an "a" - The second one, stored in
\2, contains a sequence of "p"s (in the example it contains "pp") - The third one, stored in
\3, contains the sequence "le"
Now we can print the replacement using the matches we captured: \1\2\32. Notice that we are using 3 capture values to generate "apple" and then we append a 2. This wont be interpreted as variable \32 because we can only have a total of 9 captures.
Hope this helps =)
you can first match a pattern and then change the text if matched:
echo "apple1" | sed '/apple/s/1/2/' # gives you "apple2"
this code changes 1 to 2 in all lines containing apple
This might work for you (GNU sed and Bash):
sed 's/apple1/sed "s|1|2|" <<<"&"/e' file
来源：https://stackoverflow.com/questions/12694562/sed-how-to-replace-a-character-within-a-matched-pattern-using-ampersand