Awk field sperators

[jimmy80]

I was hoping someone might be able to help - I'm trying to extract the 7th column from a file - each field is seperated by a tab - I'm using the following awk command but it doesn't return anything - i don't know has it something to do with the file sperator or what but was hoping someone else might be able to point me in the right direction!
cat testfile.txt | awk 'BEGIN {FS="\t"}{print $7}' | sort -u > test.log

Thanks,
Jimmy

 

[grega]

Enter the physical TAB character instead of \t.

cat testfile.txt | awk 'BEGIN {FS="    "}{print $7}' | sort -u > test.log

Greg.

 

[marrow]

Yes I believe tabs have affected your script. If you print $1 it works but every other $ i.e. $2 $3 will fail because of the tabs. You could substitute tabs with spaces using sed, there must be a better way but this will work.
sed 's/:/ /g' fileold > filenew

Now you can Print $7
cat filenew | awk 'BEGIN {FS=" "}{print $7}'
or
cat filenew | awk '{print $7}'

 

[PHV]

And what about this ?

t=`echo "\t\c"
awk -F"$t" '{print $7}' testfile.txt | sort -u>test.log

Hope This Help
PH.

 

[tikual]

To trace the problem, you may try my suggestion as follows:

First, see the code from the file:
sed -n l testfile.txt
Then you know the actual seperator from the file.

Second, see your seperator setting work or not.
awk -F"\t" '{print NF}' testfile.txt
Then you can know the total fields now.

In your case, I think that the problem is the tab chracter may more than one between the field. You may try:
awk '{FS = "\t*"}{print$3}' textfile.txt | sort -u > test.log
Note: The code FS = "\t*" may only work in some of version of awk. My linux is work but it isn't work in Sun. Because I found that the code is ">" rather than "\t" from the file.

tikual