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Use awk to have the fourth column with spaces

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# 1  
Old 07-16-2009
Use awk to have the fourth column with spaces
Hi Gurus,
We have a ftpserver from which we do a dir command and output it to a local file.
The content of the ftpfile is:
Code:
07-15-09  06:06AM                 5466 ABC_123_ER19057320090714082723.ZIP
07-15-09  06:07AM                 3801 ABC_123_ER19155920090714082842.ZIP
07-15-09  06:07AM                 2034 ABC_123_ER19257020090714083003.ZIP
07-15-09  06:07AM                 5346 ABC_123_ER19456120090714083105.ZIP
07-15-09  06:07AM                50188 ABC_123 4507131004299717363.ZIP
07-15-09  06:07AM                10867 ABC_69 ER194561.ZIP
07-15-09  06:07AM                73183 ABC_69_ER194631.ZIP
07-15-09  06:07AM                 1576 ABC_69_ER195427.ZIP
07-15-09  06:07AM                 5880 ABC_69_ER195428.ZIP

I generally use awk '{print $4}' ftpfile. But, I realized that the filename in the ftp might contain spaces.
So I came up with the command:
Code:
awk '{i=4;while (NF>=i) {print $i;i++}}' ftplist

But the above command prints new line for every print I give. I want it in the same line. I tried using cut but it doesn't help.
On an ad-hoc basis, I am currently using the below command:
awk '{if (NF!=4) {print $4,$NF}else {print $4}}' ftplist
The above command prints what I want but, I assume here that the filename will have only one space.
Please suggest.
# 2  
Old 07-16-2009
Try:
Code:
awk '{i=4;while (NF>=i) {printf("%s ", $i);i++};printf "\n"}' ftpfile

# 3  
Old 07-16-2009
Or:

Code:
awk 'sub(/[^ ]* *[^ ]* *[^ ]* */,"")' ftpfile

If you have GNU awk (gawk) you can avoid the repeating pattern using the --re-interval option.

And, of course, you can use sed for this.
# 4  
Old 07-16-2009
Or plain with awk:

Code:
awk '{$1=$2=$3="";sub("   ","")}1' file

Regards
# 5  
Old 07-16-2009
Thanks
Thanks that command helped. I have one more doubt. I am unable to use shell variables in awk. Could you help me again with that? I have a sample file like this:
Code:
PROJECT PATH
-----------------
ABC  /ABC/Datafiles/FILES
CLNT /CLNT/DataFiles/FILES 
BHL    /BHL/MISC/FILES
JMH /JMH/DataFiles/FILES
SMS /CLA/Datafiles/FILES
PCS /CLA/Datafiles/FILES

Now I have a script where I allow the user to enter the name of project based on which I grep the file and give the path. But here too, I realized that I am doing grep on the entire file rather than on only first column. The problem for that is the user can give anything on the given line and it might result in undesirable output.

My desired output is if the user gives PCS, he should see /CLA/DATAFILES/FILES

I initially used :
grep "$a" filename|awk '{print $2}'
where a is the name of the project which user gives.
I tried to use:
awk '{if ($1=$a) {print $2}else{print "WRONG INPUT"}}' filename

But the above awk doesn't work. Please help
Thanks
# 6  
Old 07-16-2009
You could do something like:

Code:
awk 'BEGIN {
  printf "Enter project name: "
  getline name < "-"
}
$1==name {print $2}' file

# 7  
Old 07-16-2009
Thanks
I found one more alternative:
awk '{if ( $1== "'"$a"'") {print $3}}' filename

Enclose the shell variable with (double quotes, single quotes,double quotes)

Your script also works..

Thank you.
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