Unix bash/ksh : Selection of first non space character from file from specific line

Question

I have file file1.txt whose contents are as follows:

Date List
-----------
    Quarter Date
         Year Date
             Month Date

Now I want to read the non space elements from each row of file and to write to a variable. For example for row 2 variable should contain Quarter Year only after removing space.

I tried:

tail -1 file1.txt > variable1

But it doesn't work.

Answer 1

Using sed:

variable1="$(< inputfile sed -n '3s/ *//p')"

• variable1="$([...])": runs the command [...] in a subshell and assigns its output to the variable $variable
• < inputfile: redirects the content of inputfile to sed's stdin
• -n: suppresses output

sed command breakdown:

• 3: asserts to perform the following command only on the 3rd line of input
• s: asserts to perform a substitution
• /: starts the search pattern
• *: matches zero or more characters
• /: stops the search pattern / starts the replacement string
• /: stops the replacement string (hence actually replacing with nothing) / starts the modifiers
• p: prints only the lines where the substitution succeeded

Answer 2

First read the desired line into a variable (line 3 in the example):

var=$(sed -n '3p' file1.txt)

The sed command prints (p) the 3rd line of the file. The strip the leading spaces using parameter substitution:

echo "${var#"${var%%[![:space:]]*}"}"

The inner substitution means to remove everyting except the leading spaces. The outer substitution remove those spaces at the beginning of the line.

Output is:

Quarter Date

Answer 3

tail -1 file1.txt > variable1 writes to the file variable1.

Use command substitution (bash.info 3.5.4, POSIX sh) instead:

variable1="$(tail -1 file1.txt)"

Btw my version of tail from GNU in cygwin doesn't have the -1 option. Instead, I use sed:

# EREGEX: Replace all whitespace at beginning of line
# NOTE: BSD sed uses a different flag to enable EREGEX, -E.
# EDIT: Dropped -r. \s is already included in BRE.
#       Thanks to kos for pointing that out.
# EDIT: Use POSIX [:space:] instead of Perl \s.
variable1="$(sed -e 's/^[[:space:]]*//g' < file1.txt)"

Combined with line selection:

# EDIT: limit the [s]ubstitude operation to the 4th line only, and
#       [p]rint directly from s.
variable1="$(sed -ne '4s/^[[:space:]]*//p' < file1.txt)"

Answer 4

With ksh/zsh/bash:

IFS=' ' read -r variable < <(tail -n 1 file)

read strips leading and trailing space characters if space is found in IFS (which it is by default along with tab and newline).

You can also do:

while IFS=' ' read -r variable <&3; do
  something with "$variable"
done 3< file

To process the file line by line (though that's not usually the way to go in shells) with $variable holding the current line's content with leading and trailing space characters removed.