Use grep with regular expression to extract specific substring (-o, –only-matching)

By neokrates, written on July 7, 2010

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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:

⭐ 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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2 Responses to “Use grep with regular expression to extract specific substring (-o, –only-matching)”

  1. blake says:

    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….

    • admin says:

      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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