Use space as a delimiter with cut command

You can also say:

cut -d\  -f 2

Note that there are two spaces after the backslash.

^{To complement the existing, helpful answers; tip of the hat to QZ Support for encouraging me to post a separate answer:}

Two distinct mechanisms come into play here:

• (a) whether cut itself requires the delimiter (space, in this case) passed to the -d option to be a separate argument or whether it's acceptable to append it directly to -d.

• (b) how the shell generally parses arguments before passing them to the command being invoked.

(a) is answered by a quote from the POSIX guidelines for utilities (emphasis mine)

If the SYNOPSIS of a standard utility shows an option with a mandatory option-argument [...] a conforming application shall use separate arguments for that option and its option-argument. However, a conforming implementation shall also permit applications to specify the option and option-argument in the same argument string without intervening characters.

In other words: In this case, because -d's option-argument is mandatory, you can choose whether to specify the delimiter as:

• (s) EITHER: a separate argument
• (d) OR: as a value directly attached to -d.

Once you've chosen (s) or (d), it is the shell's string-literal parsing - (b) - that matters:

• With approach (s), all of the following forms are EQUIVALENT:

  • -d ' '
  • -d " "
  • -d \<space> # <space> used to represent an actual space for technical reasons

• With approach (d), all of the following forms are EQUIVALENT:

  • -d' '
  • -d" "
  • "-d "
  • '-d '
  • d\<space>

The equivalence is explained by the shell's string-literal processing:

All solutions above result in the exact same string (in each group) by the time cut sees them:

• (s): cut sees -d, as its own argument, followed by a separate argument that contains a space char - without quotes or \ prefix!.

• (d): cut sees -d plus a space char - without quotes or \ prefix! - as part of the same argument.

The reason the forms in the respective groups are ultimately identical is twofold, based on how the shell parses string literals:

• The shell allows literal to be specified as is through a mechanism called quoting, which can take several forms:
  • single-quoted strings: the contents inside '...' is taken literally and forms a single argument
  • double-quoted strings: the contents inside "..." also forms a single argument, but is subject to interpolation (expands variable references such as $var, command substitutions ($(...) or `...`), or arithmetic expansions ($(( ... ))).
  • \-quoting of individual characters: a \ preceding a single character causes that character to be interpreted as a literal.
• Quoting is complemented by quote removal, which means that once the shell has parsed a command line, it removes the quote characters from the arguments (enclosing '...' or "..." or \ instances) - thus, the command being invoked never sees the quote characters.

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}'

cut -d ' ' -f 2

Where 2 is the field number of the space-delimited field you want.