cut

given input echo 1,2,3,4,5,6,7,8,9,...100 If I want to cut columns 5 I can do cut -d, -f-4,6- what if I want to cut multiple non consecutive columns like 5, 7,etc is there a one liner? newfurniturey You should be able to continue the sequences directly in your existing -f specification. To skip both 5 and 7, try: cut -d, -f-4,6-6,8- As you're skipping a single sequential column, this can also be written as: cut -d, -f-4,6,8- To keep it going, if you wanted to skip 5, 7, and 11, you would use: cut -d, -f-4,6-6,8-10,12- To put it into a more-clear perspective, it is easier to visualize when you

I want to use space as a delimiter with the cut command. What syntax can I use for this? cut -d ' ' -f 2 Where 2 is the field number of the space-delimited field you want. Usually if you use space as delimiter, you want to treat multiple spaces as one, because you parse the output of a command aligning some columns with spaces. (and the google search for that lead me here) In this case a single cut command is not sufficient, and you need to use: tr -s ' ' | cut -d ' ' -f 2 Or awk '{print $2}' mklement0 To complement the existing, helpful answers; tip of the hat to QZ Support for encouraging me

Without using sed or awk , only cut , how do I get the last field when the number of fields are unknown or change with every line? zedfoxus You could try something like this: echo 'maps.google.com' | rev | cut -d'.' -f 1 | rev Explanation maps.google.com's reverse will be moc.elgoog.spam cut uses dot as the delimiter and chooses the first field, which is moc lastly, we reverse it again (thanks for the reminder, @tom) to get com Use a parameter expansion. This is much more efficient than any kind of external command, cut (or grep ) included. data=foo,bar,baz,qux last=${data##*,} See BashFAQ