Shell Programming and Scripting

Solaris /bin/awk -> oawk does not allow a ! operator on a number.

Code:

awk -F\| '0==L[$1]++'

Hi,
My .psv file is getting bigger as more columns are being added. I remove duplicates based on the last column number and currently i know the position of this which is 176 . The column name is 'auditid'. Is there a way i can find the column number of this field and assign it to the array ? .


nawk -F\| '!L[$176]++' file

Assuming the column name is a field in row 1, try

Code:

awk -F\|  'NR==1 {for (i=1; i<=NF; i++) if ($i==COLNAME){COL=i; break}} !L[$COL]++' COLNAME="auditid" file

You can use the column name as a constant for the comparison, or, like here, pass it i a variable.

Just wondering... what is the logic behind using L[$1]++?

L is just an arbitrary Name for an array - call it what you like, Joe, Mimi, or L (short for logical). The $1 (first field of each respective row/line) is the index into that array, and that indexed element is incremented by ++. Any value except 0 (or empty, which is equivalent) will make the reference true, its inversion (by !) false. As the default action is print, the entire command reads: Get the array element for index $1. If it does not exist (=first occurrence of this index) invert to true, print. If it does (and has a value) , invert to false, don't print. Increment it for later print suppression.

Thank you, RudiC for the explanation. It was really helpful