Shell Programming and Scripting

Note that this is making the unstated assumption that each input line always contains exactly 15 fields. The original code seemed to be trying to end up with 15 fields no matter how many were present in the input.

The awk script here moves the fifteenth (possibly empty) field to the start of output line, but may end up with one or more output fields depending on the number of input fields.

It also assumes that decrementing NF will modify $0. Perhaps most (all?) implementations behave this way, but there are no guarantees.

Regards,
Alister

I know that decrementing NF does NOT modify $0 in awk on OS X.

I agree that decreasing NF is not guaranteed to work with every awk (for example it works with nawk on Solaris, but not with /usr/xpg4/bin/awk) ,
however, it does seem to modify $0 in my version of awk on OSX (bwk version 20070501)

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IF there are exactly 15 columns, one could try:

Code:

awk -F, '{$1=$15 FS $1; $15=x; sub(/,$/, x)}1' OFS=, file

or

Code:

awk -F, '{f=$15; $15=x; sub(/,$/, x); print f,$0}' OFS=, file

---------edit---------

This could work if there are more than 15 fields:

Code:

awk -F, '{$1=$15 FS $1; sub("(,[^,]*){" NF-14 "}$", x)}1' OFS=, file

if your awk can use interval expressions, so it will not work with mawk and (non-POSIX nawk) and gawk 3.x or lower needs gawk --re-interval