How to escape “.” with sed or grep?

Question

I have file with

denm.xyz
denm.abc
denm.def
denm_xyz
denm_abc
denm_def

I want to extract the ones with . in it.

I tried

grep "denm\.*" file
grep 'denm\.*' file

sed "/denm\.*/p" file
sed '/denm\.*/p' file

These things are matching everything in the file

However using awk

awk '/denm\./' file 

worked!!

How do I do same using grep or sed

Answer 1

You were close, you just need to remove * which means zero or more match of the preceding token, . in this case. As a result the demn_ is also being showed in the result as it matched the condition of zero ..

So you can do:

grep 'denm\.' file.txt 

Similarly in sed:

sed -n '/denm\./p' file.txt

Note that you have missed the -n option of sed, without it sed will print all the lines too.

There are also so many level of precision can be added to get exactly what you want in complex cases but take it as a start.

Example:

% grep 'denm\.' file.txt
denm.xyz
denm.abc
denm.def

% sed -n '/denm\./p' file.txt
denm.xyz
denm.abc
denm.def

Answer 2

\.* means "any number of occurrences of the character .", including zero. Since your underscored filenames start with denm followed by zero occurrences of the character ., they are matched. grep "denm\." file will work.

Answer 3

If all you need are the lines that have a . in them, used grep with fixed strings instead of regular expressions:

grep -F . file

From man grep:

-F, --fixed-strings
      Interpret PATTERN as a  list  of  fixed  strings,  separated  by
      newlines,  any  of  which is to be matched.  (-F is specified by
      POSIX.)

With fixed strings, all regular expression special characters lose their meaning.

sed does not have a corresponding option.

Answer 4

I tend to use brackets instead of backslashes for escaping:

egrep "denm[.].*" file
sed -r 's/denm[.]/.../'

This saves me the effort of thinking if and how I should escape the backslashes. (The final .* is redundant here.)