Shell Programming and Scripting

I have test string value , something like the one below

Code:

str='KUAMRJIT|GHOSH'

If I type

Code:

echo $str | grep -o -e '\|+'

it doesnt give me anything .
But on the contrary

Code:

echo $str | grep -o -e '|'

display the only one pipe character(|) thats there in the string above .
The way I understood Unix regexp and grep command is
+ : matches one / more occurence of the preceeding character
-o : displays all the occurences of patter in the file / standard input

So why the first pattern yeilds nothing but the second does ?
Does grep -o option match literal values only ?
Please help .

Thanks
Kumarjit.

You need to escape the '+', not the pipe character:

Code:

echo $str | grep -o -e '|\+'

Single quoting supersedes escaping, so it is looking for \|+ or |\+verbatim. Put the expression in double quotes. Interpreting the + needs extended regular expressions ("ERE"), so use grep -E. As | is the branch separator in ERE, you need to escape it. The -e option is dispensabe. So the following will work:

Code:

$ echo $str | grep -Eo  "\|+"
|

@Rudi, the double quotes or single quotes do not make a difference in this case:

Code:

$ echo "$str" |  grep -Eo '\|+'
|

Rats! You're right. Writing a thesis and then the basics are wrong... still it needs the -E

@Subbeh : Tried echo $str | grep -o -e '|\+' , works good , but fails while using the following command
echo $str | grep -o -e '|\*'

@kumarjt, Subbeh, that would only work with GNU grep (and then -e would be unnecessary and \+ would be a GNU extension on Basic Regular Expression and * would not need a \ )
IMO it is better to use RudiC's suggestion and use the -E option (Extended Regular Expression)