[SOLVED] get the full string when a word matches

[rudra-b]

I have files that contain line like:

(2010) 4287-4293. doi:10.1016/j.physb.2010.07.028

etc.
where the doi:* is of importance to me.
is there any way in C/shell script to get the string(even from the middle of line) begins with doi: and delimited only by space or newline?
so that from the string, I will get

10.1016/j.physb.2010.07.028

I cannot do awk(in my knowledge), as in which column of the line doi: will appear is uncertain.

[marko]

I don't know of how to get a c-shell script to do that because I don't know of a way to get regex support into a c-script but I hacked up an example perl script that works for me (below). Assuming the input file is "infile.fil", this reads the whole thing into an array, then processes it line by line from the array

#!/usr/bin/perl -w
use strict;
use warnings;

open(MF, "<infile.fil");
my @slurp = <MF>;
close(MF);
foreach my $line (@slurp) {
    if ($line =~ m/\s+doi:(.*)$/) {
        print "$1\n";
    }
}

If the input file is really huge, you'd want to not use a slurp style processing of the file but do it line by line.

The \s+ means that any amount of leading white spaces in front of the "doi:" is allowed and matches. But it requires at least one char of space. \s* would do 0 or more white spaces.

[jpollard]

I would use perl

$ perl -ne 'if (/doi:(.*)\s/){ print $1,"\n";}' <input_file >outputfile

Its shorter. Basic explaination - it only prints if the pattern is found. The pattern /doi.*)\s/ starts and ends with the "/" pattern matching string. What the pattern starts at is doi: followed by any number of characters (the .*) and ends with a whitespace (\s for spacer) which includes the end of line possibility. The "(.*)" identifies the part of the pattern (when identified) to extract - each () would identify another substring, each numbered starting from 1. So the print $1 prints the first identified substring...

This does the same as the previous poster, but uses the perl options -ne (the n puts the -e parameter in a loop that doesn't echo the input data, without the n you get the pattern identified and the following line is what it came from).

[RupertPupkin]

Originally Posted by rudra-b

I cannot do awk(in my knowledge), as in which column of the line doi: will appear is uncertain.

You could use awk's gsub function. For example, suppose you have this input file (infile.txt):

some stuff
(2010) 4287-4293. doi:10.1016/j.physb.2010.07.028
(2010) 4287-4293. doi:10.1016/j.physb.2010.07.029
some other stuff
Hey doi:!
(2010) 4287-4293. doi:10.1016/j.physb.2010.07.030
some more stuff

Then you could do this:

$ awk '$0 ~ /doi:/ {gsub(/^.*doi:/, ""); print}' infile.txt
10.1016/j.physb.2010.07.028
10.1016/j.physb.2010.07.029
!
10.1016/j.physb.2010.07.030

[stevea]

grep -o "doi:.*" /tmp/foo | cut -b5-

[ocratato]

Most of the answers seem to haved missed " and delimited only by space or newline?" which leads me to think that there might sometimes be more stuff on the end of the line occasionally.

sed -n 's/\(^.*\)doi:\([^ ]*\)\(.*$\)/\2/p' < infile > outfile

should do the job.

[rudra-b]

Thanks to all of you. ocratato's solution worked.