Using sed to cut a string starting with n spaces and ending with n spaces

Question

How to use sed to cut text starting with x spaces and ending with y spaces?

For instance This is my string:

 kkk 111 fff      aaabbb 5d98 ccc         mmmppp 9369d

and I want to get this output:

 aaabbb 5d98 ccc

(the number of spaces is not known)

Thank you.

Answer 1

We told about some text staring with unknown quantity of spaces so

sed 's/.* \{2,\}\([[:alnum:]].*\) \{2,\}.*/\1/'

or with -r (-E)

sed -E 's/.* {2,}([[:alnum:]].*) {2,}.*/\1/'

seems to enough but grep is better in the case

grep -Po ' {2,}\K[[:alnum:]].*(?= {2,})'

And not so strong (just with two whitespaces) but correct too:

sed -E 's/.*  (\w.*)  .*/\1/'

Answer 2

You can use -r option for extended regular expression where number of characters can be specified inside {}, so the following will print all words surrounded by 6 spaces:

sed -r 's/.* {6}(\w*) {6}.*/\1/'

In case if the title has spaces too, the better choice would be

sed -r 's/.* {6}(.*) {6}.*/\1/'

Answer 3

Edit: I've borrowed the -r flag (enables extended regex syntax) from jimmij to cure backlashitis.

The following works, under the following conditions:

• you are willing to say that the field separator is at least n spaces, e.g. 3
• the contents of the field of interest do not include a space anywhere.

In that case, this regex works:

    echo ' 01      Title      Chapter 01' |
    sed -r 's/^.* {3,}([^ ]+) {3,}.*$/\1/'

Or, in case you like your backslashes, this is what this looks like in non-extended regex syntax:

    echo ' 01      Title      Chapter 01' |
    sed 's/^.* \{3,\}\([^ ]\+\) \{3,\}.*$/\1/'

Explanation of the regex:

^        start of line
.*       any number of characters at the start of the line
 {3,}    at least 3 spaces
([^ ]+)  1 or more non-space characters (capture this group as \1)
 {3,}    at least 3 spaces
.*       anything on the rest of the line
$        end of the line. Not needed, because of the .*, but nicely explicit.

Answer 4

Assuming you want the same number of spaces on either side:

$ sed -r 's/(^|.*[^[:space:]])([[:space:]]+)([^[:space:]]+)\2([^[:space:]].*|$)/\3/g' <<<"01      Title      Chapter 01"
Title

(I used the character class instead of just , with just a space, the expression should be considerably shorter: sed -r 's/(^|.*[^ ])( +)([^ ]+)\2([^ ].*|$)/\3/g').

By using the backreference within LHS, we ensure that the same number of spaces are present on both sides.

Answer 5

I believe you are trying to catch the title ?

here us a way to catch things by getting rid of the first word, and the last 2 words, and displaying the rest (spaces included) :

awk '{ $1=""; $(NF-1)="" ; $NF="" ; print $0}'

Or even better: get rid of first element, and discard last 2, and also the extra spaces ( changing a $n, or NF, forces a redraw of $0 on most awk implementation):

awk '{ shift ; NF=(NF-2); print $0}'

example

$   echo   ' 01      Title is   here!     Chapter 01' | awk '{ shift ; NF=(NF-2); print $0}'

 Title is here!

The advantage of awk is that it is easy to add tests (is $1 an integer? is $(NF-1) "Chapter" ? etc)