How to specify more spaces for the delimiter using cut?

Question

Is there any way to specify a field delimiter for more spaces with the cut command? (like " "+) ? For example: In the following string, I like to reach value '3744', what field delimiter I should say?

$ps axu | grep jboss

jboss     2574  0.0  0.0   3744  1092 ?        S    Aug17   0:00 /bin/sh /usr/java/jboss/bin/run.sh -c example.com -b 0.0.0.0

cut -d' ' is not what I want, for it's only for one single space. awk is not what I am looking for either, but how to do with 'cut'?

thanks.

Answer 1

Actually awk is exactly the tool you should be looking into:

ps axu | grep '[j]boss' | awk '{print $5}'

or you can ditch the grep altogether since awk knows about regular expressions:

ps axu | awk '/[j]boss/ {print $5}'

But if, for some bizarre reason, you really can't use awk, there are other simpler things you can do, like collapse all whitespace to a single space first:

ps axu | grep '[j]boss' | sed 's/\s\s*/ /g' | cut -d' ' -f5

---

That grep trick, by the way, is a neat way to only get the jboss processes and not the grep jboss one (ditto for the awk variant as well).

The grep process will have a literal grep [j]boss in its process command so will not be caught by the grep itself, which is looking for the character class [j] followed by boss.

This is a nifty way to avoid the | grep xyz | grep -v grep paradigm that some people use.

Answer 2

awk version is probably the best way to go, but you can also use cut if you firstly squeeze the repeats with tr:

ps axu | grep jbos[s] | tr -s ' ' | cut -d' ' -f5
#        ^^^^^^^^^^^^   ^^^^^^^^^   ^^^^^^^^^^^^^
#              |            |             |
#              |            |       get 5th field
#              |            |
#              |        squeeze spaces
#              |
#        avoid grep itself to appear in the list