UNIX for Dummies Questions & Answers

Hello,

I have a file in which there are many lines.
I don't know the length of any line and they also vary.
I want to pick the last word of those lines which end with for eg. xyz

Thanks in Advance

i'm not sure what the above meant by ...

but , to get the last word of all lines in file
do the following

Code:

awk '{ print $NF }' file1

If I understand the problem, I think it's more like this...

Code:

sed -e :a -e 's/\(.*\)\( .*xyz\)$/\1/;ta' -e's/.*xyz$//;s/.* //'

Hello bhargav,

awk '{ print $NF }' file1
worked out
Thanks

Perderabo:

Your script gave me the last word of the last line, but i require the last word of each line ending with xyz.
and also i am quite novice to 'sed' programming and would request you to please explain that script to me.

From google:


awk's basic mode of operation is to read its input, chop each line into fields separated by some delimiter (white space by default, but you can change it), and then allow you to do pattern matching and other operations based on those fields. The thing I use it for most often is to grab a particular field.

Let's look at a trusty long ls listing:


--------------------------------------------------------------------------------

% ls -l
-rw-r--r-- 1 jeffy 28 May 9 16:12 Makefile
-rwxr-xr-x 1 jeffy 24576 May 28 11:31 foo
-rw-r--r-- 1 jeffy 57 May 9 16:13 foo.c
-rw-r--r-- 1 jeffy 57 May 28 11:37 foobar
-rw-r--r-- 1 jeffy 71 Jun 2 11:45 fumpty


--------------------------------------------------------------------------------
Suppose I want to grab just the file sizes for some reason. awk numbers fields starting with 1 (not zero like you'd expect from a bunch of unix geeks), so we count across and see that we want to print out field 4, so just do this:
--------------------------------------------------------------------------------

% ls -l | awk '{print $4}'
28
24576
57
57
71


--------------------------------------------------------------------------------
Easy as pie. Notice that the awk program is enclosed in single quotes. This protects the "$4" from the shell so it gets evaluated by awk, not csh (or whatever)
You can print out multiple columns in any order by separating them with commas:


--------------------------------------------------------------------------------

% ls -l | awk '{print $3, $1, $4, $NF}'
jeffy -rw-r--r-- 28 Makefile
jeffy -rwxr-xr-x 24576 foo
jeffy -rw-r--r-- 57 foo.c
jeffy -rw-r--r-- 57 foobar
jeffy -rw-r--r-- 71 fumpty


--------------------------------------------------------------------------------
Notice that the separating white space is not preserved, but gets scrunched down to a single space.
Wait a minute, what's with that "$NF" in that last example? NF is an internal awk variable that always represents the Number of Fields in the current line. By sticking a dollar sign in front of it, I get the equivalent of a "$8" when I run the script on the "ls -l" output. But I don't have to know how many fields there are, I can just grab the last one.


-Thanks .--from your question - it made me learn "awk"!!

One more thing: field seperator

awk -F:

the "-F" flag sets the field separator character as :

white space by default, separating white space is not preserved, but gets scrunched down to a single space.

hope this hekps!!

Thanks,

In awk, we have a BEGIN and END construct.
Generally we place initialisations in BEGIN and final code in END.
( See i have started it also ).

My basic awk script is:

Begin {print ...} /regurlar-exp/ file End {print ...}

Its working fine.
But sometimes, the file which i am using as input becomes empty because its generated by some other script file.
Now only my End statements get printed, the Begin statements get bypassed.

Can somebody help me as to why is this happening and how can i solve it.