awk: How can I return a specified record
 

Hi,

In awk , we choose which field to return by variables $1,$2 ,......

this will list the fields wanted for all records (where record is a single line of the returned values), the returned values may have more that 1 record usually.

I need to select a single record , mean to get the output of the awk result line by line.

how can this be doen ,

( I need this for many things , ex: deleting all directories with name ".sss" (by filtering the find results through awk then pass record-by-record to rm) , and may others usages.


Thanks.

Hi,

Hope I understand correctly, but is this what you want/need:

awk '/root/ { print $0 }' /etc/passwd
This only prints lines that contain root from the /etc/passwd file.

You could also check a specific field and print another field if a hit is found:

awk -F":" '$3 == "0" {print $1}' /etc/passwd
Print first field (login name) if third field (UID) equals zero.

Hope this helps.

Thanks for your replay ,, let me give an example please:
I wonder how to delete them ? i have thought in awk , I want awk to return one line (RECORD) at a time so i can delete them through redirecting (awk or other) output to rm.

also:
and I need to kill them all, so I need to get the values line-by-line or in awk terms: as records.

Thanks.

The 2 easiest ways are:

Use the -exec option that comes with find. I.e:
find ./ -name ".svn" -exec rm {} \;
If you want to know, beforehand, what will be deleted change rm to ls.

Another way is:

find ./ -name ".svn" | xargs rm

The find command, in both examples, will return what you are looking for and gives this to (example one) rm by using the -exec parameter, or pipes it to xargs rm in the second example.

Here you can pipe it to xargs rm. I.e:

ps -aux | awk '/httpd/ {print $2}' | xargs rm

man xargs for some gory details.

Hope this helps.

Many thanks for the new command (for me) xargs ...