awk

[KT_4]

I am using awk to extract the 2nd column (a list of file names) out of a file like this:

07:37:46 file1.txt 02/12/06
07:35:48 file 2.txt 02/12/06
07:36:55 file3.txt 02/12/06

If you notice the 2nd record has a space as part of the file name so awk is treating this as a field separator.. Is there a way to get awk to include the space so I get the entire file name extracted in the 2nd record?

[Robert Bennett_3 · Accepted solution]

try - awk '{print $2,$3}' infile > outfile

[Indira Aramandla]

Hi KT,

Since you text file has lines with common words "txt" and "file",

you can change your file to do a global replacement with a comma in front of file and after txt. To modify vi the file and do this
:1,$s/file/,file/g and then save the file
Then again :1,$s/txt/txt,/g and save the file.

Eg:-
Original file conntents
07:37:46 file1.txt 02/12/06
07:35:48 file 2.txt 02/12/06
07:36:55 file3.txt 02/12/06

The modified file contents
07:37:46 ,file1.txt, 02/12/06
07:35:48 ,file 2.txt, 02/12/06
07:36:55 ,file3.txt, 02/12/06

Now you can use awk to list he files names as follows.
awk -F, '{ print $2 }' filename

This will output the file names.
file1.txt
file 2.txt
file3.txt


Indira A

[Muthukumar_5]

hai,

we can do as,

# cat > testfile
07:37:46 file1.txt 02/12/06
07:35:48 file 2.txt 02/12/06
07:36:55 file3.txt 02/12/06

# awk '{ if (NF==4) { print $2$3 } else { print $2 } }' testfile
file1.txt
file2.txt
file3.txt

it will check that if a input line is containing 4 fields then it will add 2nd and 3rd. Else it will display only 2nd.

HTH.

[Thierry Poels_1]

how about a cryptic sed :)

sed 's;\(..:..:.. \)\(.*\)\(../../..\)$;\2;' yourfile

regards,
Thierry.

[Jdamian]

Hi.

I prefer to think a little before answer. I assume the following:

a) first field is ALWAYS a timestamp in the format HH:MM:SS
b) last field is ALWAYS a date in the format xx/yy/zz...
then first and last fields won't contain blanks.

c) everything between first and last fields is a file name.

My first proposal would be:

awk '{ NAME=""; for(i=2; i
It works fine if blank chars in file name are single (i.e. not more than one blank) and file name doesn't begin nor end in blank chars -- I assume that JUST a blank char is used as separator in the original line.

A better proposal will remove the first field (plus a blank char after it) and last field (plus a blank char befor it) from the original line... leaving alone the file name.

[Keith Bryson]

Assuming your file is in /tmp/testfile:

cat /tmp/testfile | while read a
do
templength=`echo $a | wc -m`
let length=templength-10
echo $a | cut -c10-$length
done

This assumes that you always have a time record at the start and date record at the end (both are always the same length and the script strips the front and end 10 characters).

All the best - Keith

[Jdamian]

my ultimate proposal is:

awk '
{
LINE=$0 # needed to preserve original line and fields.
sub("^[[:blank:]]*" $1 ".", "", LINE) # remove first field plus a separator char

sub("." $NF "[[:blank:]]*$", "", LINE) # remove a separator char plus last field

print LINE
}'

one doubt: I'm not sure if char class [[:blank:]] is more suitable than [[:space:]].

[Gordon Morrison]

Why be AWKward? I have always found AWK syntax to be tortuously finnicky and impossible to remember, and I have always found that there is another way (usually built-in to ksh) that does the same thing in a much easier & friendlier way:
read a line from your file into a variable (in this case, ${line}) then let ksh do it:

temp=${line#* }
fname=${temp% *}

This works whether or not there is a space in the filename on $line

[Thierry Poels_1]

yep, nice one.

while read line
do
temp=${line#* }
fname=${temp% *}
echo $fname
done < tt

10 points to Gordon.