Use grep with regular expression to extract specific substring (-o, –only-matching)By neokrates, written on July 7, 2010 |
 bash snippet |
- neokrates
- Email: uwarov@yahoo.com
- Website: http://www.thinkplexx.com
- Join date: 05-31-09
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I have a file with some random text and I wish a very specific part of it to be assigned to my bash variable.
File:
$ cat grepme.txt My name is &&&neokrates3223 My name is neokrates-- My name is tadada Ma name is not neokrites
And I wish to know the name.
Suppose, the correct name can be identified:
![[star]](http://www.thinkplexx.com/wp-includes/images/smilies/thinkplexx/set1/StarX16.png)
it is prefixed by “My name is”;
it is surrounded by &&& and some numbers;
I use some grep with pipe to extract the name:
MY_NAME=$(grep -o "My name is [\&\/a-zA-Z0-9\-]*" grepme.txt |grep -o "&[\&\/a-zA-Z0-9\-]*" | grep -o "[a-zA-Z]*" ; echo -e "\n\nYour name is $MY_NAME\n\n"
I get:
Your name is neokrates
Have fun 
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there is no closing bracket. also im fairly sure the output of grep -o cannot be piped to another grep – at least its not working for me….
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Hi Blake,
you are right about bracket, publishing problem I guess.
Grep can be generally piped. What Version of grep and under which Os fails to pipe “grep -o“?
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