How to escape regular expressions with grep ?

Hi all,

I want to find the number of occurances of each word in a given file.

The following script works fine for normal words.


However
when there is a word like ^$ in the file grep takes that word as a regular expression. How can I avoid this ?

I cannot escape ^$ to \^\$ since there may be so many regular expressions (of different forms ) in the file.

-Thanks
Murugesan

I had a similar problem with the $ characters in words. I was not able to escape them and gave up and have written my own version of limited grep in C.

Perhaps somebody knows a better way.

Regards

I had a similar problem. I was trying to grep -v for a filename that began with a $. (still don't know how that one got created!) I found this and it seems to work for me (ksh on Solaris 9):

grep -v '\$'

Hope that helps!

we try to write unix shell

use
for i in `tr ' ' '\n' dinesh_p_v_FileName¦sort -u`
do
printf "Word : %-20s Count : %-5d\n" $i `grep "$i" words | wc -l`

done
#be a little clear
empty=`grep -c '^$' dinesh_p_v_FileName`
printf "Blank lines count : %d\n" $empty
# i still prefer the old 'echo'
rm -f nothing

NOTA: that 'i' is also a very common programm stupidity
suppose in a 20,000 lines src code you have to change that 'i'
by 'iii'
try it using 'n' or 'c'