This is the error I am getting and it's failing because of a variable whose value is supposed to be 2 (I am getting this using a select * from tabel). I am getting spaces in that variable.

+ 0 != 
./setjobs[19]: 0:  not found.

How do I remove all those spaces or a newline from that variable? Can tr, sed, or anything help?

This what I am doing:

set_jobs_count=$(echo  "set heading off;
      select count(*) from oppar_db
      where ( oppar_db_job_name, oppar_db_job_rec ) in ($var) ;" | \

This works as suggested:

| sed 's/[[:space:]]//g'

But I still obtain a value like :

  • 1
    You can cast a string to an int in the select statement. How that is done depends on database, Sybase, Oracle, MySQL, etc. – bsd Feb 24 '12 at 13:05
  • 1
    how do i do that, i have oracle 9i – munish Feb 24 '12 at 13:11
  • 1
    using sed it's | sed 's/[[:space:]]//g' to collapse whitespace – bsd Feb 24 '12 at 14:06
  • thanks works up to some extent but still the values of variable comes like set_jobs_count= 2 – munish Feb 24 '12 at 14:37
  • Depending on the shell, you may be able to do it without any external tools. See stackoverflow.com/a/3352015/587717 – Edd Steel Feb 24 '12 at 16:30

You can use tr, as in tr -d '\040\011\012\015', which will remove spaces, tabs, carriage returns and newlines.

  • 3
    Is there an advantage of using \040\011\012\015 over [:space:]? – Nick Mar 11 '15 at 2:04
  • Portability with very old UNIX versions is the only reason I can think of--- old enough that the UNIX version predates POSIX.1. – Kyle Jones Mar 11 '15 at 19:15

In ksh, bash or zsh:


In any shell, you can remove leading and trailing whitespace and normalize all intermediate whitespace to a single space like this:

set +f
set -- $set_jobs_count
set -f

set +f turns off globbing; if you know that the data contains none of the characters \[?*, you can omit it.

  • interesting answer +1 – munish Mar 1 '12 at 15:02
  • @BinaryZebra The globbing happens at set -- $set_jobs_count. set_jobs_count=$* is equivalent to set_jobs_count="$@" since $* and $@ are only equivalent when unquoted and the right-hand side of an assignment is parsed the same way as a double-quoted string. – Gilles 'SO- stop being evil' Jul 26 '15 at 22:55

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