what can be the regex for the following string

Question

I am doing this in groovy.

Input:

hip_abc_batch   hip_ndnh_4_abc_copy_from_stgig abc_copy_from_stgig
hiv_daiv_batch  hip_a_de_copy_from_staging abc_a_de_copy_from_staging

I want to get the last column. basically anything that starts with abc_.

I tried the following regex (works for second line but not second.

\abc_.*\

but that gives me everything after abc_batch

I am looking for a regex that will fetch me anything that starts with abc_ but I can not use \^abc_.*\ since the whole string does not start with abc_

Answer 1

Try this:

/\s(abc_.*)$/m

Here is a commented version so you can understand how it works:

\s          # match one whitepace character
(abc_.*)    # capture a string that starts with "abc_" and is followed
            # by any character zero or more times
$           # match the end of the string

Since the regular expression has the "m" switch it will be a multi-line expression. This allows the $ to match the end of each line rather than the end of the entire string itself.

Edit: You don't need to trim the whitespace as the second capture group contains just the text. After a cursory scan of this tutorial I believe this is the way to grab the value of a capture group using Groovy:

matcher = (yourString =~ /\s(abc_.*)$/m)
// this is how you would extract the value from 
// the matcher object
matcher[0][1]

Answer 2

It sounds like you're looking for "words" (i.e., sequences that don't include spaces) that begin with abc_. You might try:

/\babc_.*\b/

The \b means (in some regular expression flavors) "word boundary."

Answer 3

Regex buddy(pay) and RegExr(free) can be a big help in learning RegEx if you are interested.

Answer 4

I think you are looking for this: \s(abc_[a-zA-Z_]*)$

If you are using perl and you read all lines into one string, don't forget to set the the "m" option on your regex (that stands for "Treat string as multiple lines").

Oh, and Regex Coach is your free friend.