UNIX for Dummies Questions & Answers

Hi

I have a string which can be in either the format

$string="abcd.efgh"

or

$string="abcd..efgh"

or

$string="abcd...efgh"

what I want to do is pick out any string which has more than 1 '.'
ie I want to pick out abcd..efgh and abcd...efgh but NOT abcd.efgh

I have been using the following statement

echo $string | grep '..'

This statement picks out everything INCLUDING abcd.efgh
I can't understand why this happens. Can anyone help me to solve this problem?

Many thanks!
Helen

Grep, by default will either print the given line that matches or with '-v' prevent the printing an entire line if the criteria matches.


My guess is that you are trying to keep the string with 1 . in the middle and remove any extra ..s that are in other occurences of the string. Is that right??? OR to remove the words that contain .. or ... entirely from the file.

CASE #1
***********************
Use sed -e 's/D..E/D.E/' -e 's/D...E/D.E/' file.orig > file.new


Here is from the man page. You can do multiple changes at once similar to grep -e option

sed -e 's/abc/xyz/' -e 's/lmn/rst/' file1 >file1.out


CASE #2
***********************
If you are trying to remove "pick" out the occurences that have .. or ... in the middle. you need to give

Use sed -e 's/D..E//' -e 's/D...E//' file.orig > file.new.

This will remove the offending words from the file.

I don't need to use sed but I think I've managed to work out how to do it:

I believe the answer is

echo $string | grep '/./.'

It seems to work anyway. If anyone knows of a better way, please let me know.

Thanks
Helen

The dot is a special character when using grep....

grep '\.\.' works (not sure how you got grep '/./.' to work though?

Basically the . is a special character - so to make it be taken literally you put in the backslash (and you do need it for both dots!).

Yes, I did use '\.\.' - my head was so fuzzy after all that thinking that I typed the post in wrong!