Question

Please help this problem.

I have this command:

awk 'BEGIN {printf "%-15s", "Date of birth:" > "/dev/stderr"; getline var; print "Today finished:", var ,"days"}'

How do I print ‘var’ with only the first command ‘printf’? Can I remove the second command ‘print’?

Thank you for your help.

EDIT-1:

answer for Mat

The first example is good, but prints ‘var’ in the next line. Is it possible to print the ‘var’ in the same line?

EDIT-2:

answer for Peter.O

Thank you for your help. Exactly the point. Data will be used for further calculations. Result of the calculation will be printed in the same line. Example:

awk 'BEGIN {printf "%-15s", "Date of birth:" > "/dev/stderr"; getline var; print "Today finished:", var ,"days"}'`

That I want to get the result:

Date of birth: 2011-02-23 Today finished: 2011-02-23 days

EDIT-3:

answer for shellter

Thank you for your comments.

Answer 1

In your example code, you have redirected printf to stderr. I assume that is because you don’t want "Enter the date:" going to the final output (stdout) which will be further processed … That works and is a clean way of disposing of that unwanted text.

So if you want your formatted output:
Date of birth: 2011-02-23 Today finished: 2011-02-23 days
to go to standard output (stdout), then the answer to your question is simple: You definitely need to call another instance of print/printf which is directed to standard out.

print/printf has no concept of where its output is going; it just prints … The redirection operator > has no connection whatsoever to the internal workings of print(f), so you can’t get a single call of print(f) to send some output to stderr and the rest to stdout.