How can I match . (actual dot) using sed?

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1

Hi All,
How can I match . (actual dot) using sed? Please help.

Thanks

2

You need to escape it with a \

Look at this.

sh-2.05b$ cat jing.txt
test
.
test.match

sh-2.05b$ sed -n -e '/\./p' jing.txt
.
test.match
3

Thanks vino.

But this doesn't work. I am looking for /./* (one or more of /./ and replace with a single /./

I tried dereferencing . and /
it doen't work. Please help.

Thanks

4

This is going to be a confusing regex.

sed -e 's#\/\.\/\{1,\}#\/\.\/#g'

vino

5

Please see below:

cat 1 gives:

/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./

sed -e 's#\.\{1,\}#\.#g' 1 > 2

now see cat 2

/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./

It is not working.

6

Yes the previous solution will not work.

But this should work

sh-2.05b$ cat jing.txt 
test
/./
test/.//.//./match
sh-2.05b$ sed  -e 's#\(\/\.\/\)*#\1#g' jing.txt 
test
/./
test/./match

vino

7 
cat 1
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./

sed  -e 's#\(\/\.\/\)*#\1#g' 1 > 2

cat 2
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./
/.//.//./

It is not working. Is there is special reason why this doesn't work?

8

Possibly your sed version could be the culprit in this case.

I am using this sed

sh-2.05b$ sed -V
GNU sed version 4.0.7

vino

9

Vino,
See below:

sed -V
sed: illegal option -- V

??????? :confused: :confused: :confused:

10

Not strictly speaking what you asked for, but should do what you want.

sed -e 's#/\./[./]*#/\./#g' file