Found a word

[iranzo]

Hello,

How to extract a word or a field in a file
depending on the position of another word, field ?

variable:
MODPORT="2/6"

here the text file :
2/6 ARASW25 TOTO tutu
(ARASW26) 2/6 TOTO xyz 12/x
tutu yy uu 6/8 ARASW56 titi fff

with a command like:
grep $MODPORT fic | ....

We want to extract only the words beginning
by "ARASW" with the variable "MODPORT":
ARASW25
ARASW26

Thanks a lot.(if you understand my problem)

[A. Clay Stephenson]

You have managed to confuse me and I never solve a problem until I understand it.

You said for input:
2/6 ARASW25 TOTO tutu
(ARASW26) 2/6 TOTO xyz 12/x
tutu yy uu 6/8 ARASW56 titi fff

You want:
ARASW25
ARASW26

Note that "(ARASW26)" does not start with "ARASW" but is included in the output; however, "ARASW56" does start with "ARASW" but is not included in the output.

Grep will not do this but sed, perl, or awk can easily grab the field. You should also tell us whether "ARASW" appears at most once in any given line of input. If more than once then what should the output be?
"ARASW"

[Stephen Keane]

grep $MODPORT fic | sed -e "s/^.*ARASW/ARASW/" -e "s/[^[:alnum:]].*//"

(That's all on one line by the way.)

Possibly, not sure if I understand your question fully.

[A. Clay Stephenson]

Oh, I see, you first want to find any line containing "2/6" but this is flawed as well. What about lines containing "12/6", "2/60", etc. Are the only characters which could surround your "ARASW", '(' and ')'?

[Jeff Schussele]

Hi,

grep can only handle entire lines (records) by itself. awk can handle words (fields) in addition to records
So you need to use a combination of grep & awk to do this.

Rgds,
Jeff

[James R. Ferguson]

Hi:

Perhaps this helps:

# cat /tmp/perl.pl
#!/usr/bin/perl -w
$x=shift;
$f=shift;
open( FH, "<", $f ) or die "Can't open $f: $!\n";
while () {
if (m%$x%go) {
print "$`" if defined $`;
print "$'" if defined $';
}
}

Run as:

# /tmp/perl.pl 2/6 file

...where "2/6" is your MODPORT variable value and "file" is your input.

Regards!

...JRF...

[Kent Ostby]

I'm going to post two things. The first will pull the words that CONTAIN ARASW as follows.

Create a file called pullit.awk and put the following lines in it:

/ARASW/ {idx1=index($0,"ARASW");
sbs1=substr($0,idx1);
idx2=index(sbs1," ");
if (idx2==0) {print sbs2;next;};
print substr(sbs1,1,idx2);
}

Run the command:

awk -f pullit.awk < inputfile

Output will be:

ARASW25
ARASW26)
ARASW56

[Kent Ostby]

Oops .. okay .. see that by re-reading this, I see that you also want the line to have the same value as $MODPORT.

so here is what I would do.

Create a file pullit.sh

echo $MODPORT > useme
cat textfile >> useme
awk -f pullit.awk < useme

Create another file called pullit.awk:

NR==1 {checkvalue=$0;next;}
index($0,checkvalue)==0{next}
/ARASW/ {idx1=index($0,"ARASW");
sbs1=substr($0,idx1);
idx2=index(sbs1," ");
if (idx2==0) {print sbs2;next;};
print substr(sbs1,1,idx2);
}

This will yield:
ARASW25
ARASW26)

If you know that you will haev ARASW?? then you can use this for pullit.awk instead:

NR==1 {checkvalue=$0;next;}
index($0,checkvalue)==0{next}
/ARASW/ {idx1=index($0,"ARASW");
print substr($0,idx1,7);
}

This will yield:

ARASW25
ARASW26

[Sandman!]

How about the awk construct below:

awk '$0~/2\/6/ {print substr($0,match($0,/ARASW/),7)}'

cheers!

[James R. Ferguson]

Hi Iranzo:

This gives you exactly what you described whether or not your "word" occurs before or after the 'MODPORT' variable's value. It's a generalized script to show the word that preceeds or follows the token specified:

# cat /tmp/perl.pl
#!/usr/bin/perl
use strict;
use warnings;
my ($x, $f, $part1, $part2, @flds);
$x=shift;
$f=shift;
open( FH, "<", $f ) or die "Can't open $f: $!\n";
while () {
if (m%$x%go) {
($part1 = $`) =~ s/[()]//g;
($part2 = $') =~ s/[()]//g;
if ($part1 =~ m/\s+/) {
print "$part1\n";
} else {
@flds = split / /, $part2;
print "$flds[1]\n";
}
}
}
1;

--cut--

NOTE that the "word" that occurs before or after 'MODPORT' is printed. Given your input, you will see your requested output.

As before, run as:

# MODPORT="2/6";/tmp/perl.pl ${MODPORT} filename

Regards!

...JRF...

[iranzo]

For Stephenson

We just want lines containing 2/6
Surround ARASW the only characters
are ( and )

Thanks

[Muthukumar_5]

Do you want to grep the 2/6 pattern from a file and check for ARASW string enclosed with ()??

# cat file1
2/6 ARASW25 TOTO tutu
(ARASW26) 2/6 TOTO xyz 12/x
tutu yy uu 6/8 ARASW56 titi fff
# grep "2/6" file1 | grep '^(ARASW[^ ]*)'
(ARASW26) 2/6 TOTO xyz 12/x

If it is not the requirement revert with more details.

hth.

[James R. Ferguson]

Hi Iranzo:

Please try the second solution I posted above. Based on your input, in my opinion, it *does* give you exactly the output for which you asked.

Cut-and-paste the script I posted in my *second* post, name it "perl.pl" and call it like:

# ./perl.pl 2/6 yourfile

The first argument (for example, 2/6, tells that you only want to find lines with this token in them.

Regards!

...JRF...

[Muthukumar_5]

Could not understand your requirement in the beginning:(

# grep $MODPORT fic | awk '{ for (i=1;i
hth.

[Muthukumar_5]

Simpy as,

sed -e "/2\/6/!d;s/.*\(ARASW[^) ]*\).*/\1/;"

hth.

[Sandman!]

Iranzo,

Based on your reply here's the complete awk construct...

# awk '$0~/ 2\/6 /||$0~/^2\/6 /||$0~/ 2\/6$/{print substr($0,match($0,/ARASW/),7)}'

cheers!