Print second-to-last column/field in `awk`

awk

I want to print the second-to-last column or field in awk. The number of fields is the NF variable. I know that I should be able to use $NF, but I'm not sure how it can be used.

And this does not seem to work:

awk ' { print ( $NF-- )  } '

Best Answer

awk '{print $(NF-1)}'

Should work

Related Solutions

Linux – Using awk to print all columns from the nth to the last

Print all columns:

awk '{print $0}' somefile

Print all but the first column:

awk '{$1=""; print $0}' somefile

Print all but the first two columns:

awk '{$1=$2=""; print $0}' somefile

Awk + one line awk syntax to print only once the second field if “true” word matched

Here is one way:

awk '{if ($1 == "false") {array[$2] = $1} else if (array[$2] != "false") array[$2] = $1} END {for (i in array) if (array[i] == "true") print i}' inputfile

Edit:

Here it is on multiple lines:

awk '{if ($1 == "false") 
         {array[$2] = $1} 
     else if (array[$2] != "false") 
         array[$2] = $1} 
     END {for (i in array)
             if (array[i] == "true") 
                 print i}
' inputfile

Best Answer

Related Solutions

Linux – Using awk to print all columns from the nth to the last

Awk + one line awk syntax to print only once the second field if “true” word matched

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