How to read file and only output certain content

Shell Programming and Scripting
1

Hi -

I have a file containing data like :-

cn=tommy,cn=users,c=uk
passwordexpirydate=20100530130623z

cn=jane,cn=users,c=uk
passwordexpirydate=20100423140734z

cn=michael,cn=users,c=uk
passwordexpirydate=20100331020044z

I want to end up with a file that looks like:-

cn=tommy,cn=users,c=uk
20100530130623
cn=jane,cn=users,c=uk
20100423140734
cn=michael,cn=users,c=uk
20100331020044

I can very inefficiently first remove the blank lines using something like:-

sed '/^$/d' file1 > file2

I can use awk to just extract the date without the "passwordexpirydate=" and "z".

But what I can not do is do eveything I need to do in one loop .. e.g. using a while read line do ... done loop.

Any ideas please?

2

Hi,

sed s/^passwordexpirydate=*// IN_FILE| sed '/^$/d' | sed s/z$//

It's sure there is another form in only one sentence....wait for the responses of sed and awk's gurus...

3

Try:

sed "/^$/d;s/^password[^=]*=\(.*\)z$/\1/" file

HTH Chris

4

Thanks for the replies. Both answers work perfectly.

---------- Post updated at 03:02 PM ---------- Previous update was at 12:18 PM ----------

This is suddenly got more complicated!!

I need to do what I did before but also delete any line starting cn=users or cn=Users and the following line.

So,

cn=tommy,cn=users,c=uk
passwordexpirydate=20100530130623z

cn=Users,cn=users,c=uk
passwordexpirydate=20100423100000z

cn=jane,cn=users,c=uk
passwordexpirydate=20100423140734z

cn=michael,cn=users,c=uk
passwordexpirydate=20100331020044z

would end up as :-

cn=tommy,cn=users,c=uk
20100530130623
cn=jane,cn=users,c=uk
20100423140734
cn=michael,cn=users,c=uk
20100331020044

Thanks in advance for any help.

5 
sed "/^$/d;/^cn=Users/{N;d;};s/^password[^=]*=\(.*\)z$/\1/" file
6

Also,

awk -F= '/^cn/ {print} /^passwordexpirydate/ { gsub("z","",$2);print $2}' file
7

Cheers thanks all for the help.