How to make the 'cut' command treat same sequental delimiters as one?

As you comment in your question, awk is really the way to go. To use cut is possible together with tr -s to squeeze spaces, as kev's answer shows.

Let me however go through all the possible combinations for future readers. Explanations are at the Test section.

tr | cut

tr -s ' ' < file | cut -d' ' -f4

awk

awk '{print $4}' file

bash

while read -r _ _ _ myfield _
do
   echo "forth field: $myfield"
done < file

sed

sed -r 's/^([^ ]*[ ]*){3}([^ ]*).*/\2/' file

---

Tests

Given this file, let's test the commands:

$ cat a
this   is    line     1 more text
this      is line    2     more text
this    is line 3     more text
this is   line 4            more    text

tr | cut

$ cut -d' ' -f4 a
is
                        # it does not show what we want!


$ tr -s ' ' < a | cut -d' ' -f4
1
2                       # this makes it!
3
4
$

awk

$ awk '{print $4}' a
1
2
3
4

bash

This reads the fields sequentially. By using _ we indicate that this is a throwaway variable as a "junk variable" to ignore these fields. This way, we store $myfield as the 4th field in the file, no matter the spaces in between them.

$ while read -r _ _ _ a _; do echo "4th field: $a"; done < a
4th field: 1
4th field: 2
4th field: 3
4th field: 4

sed

This catches three groups of spaces and no spaces with ([^ ]*[ ]*){3}. Then, it catches whatever coming until a space as the 4th field, that it is finally printed with \1.

$ sed -r 's/^([^ ]*[ ]*){3}([^ ]*).*/\2/' a
1
2
3
4

Try:

tr -s ' ' <text.txt | cut -d ' ' -f4

From the tr man page:

-s, --squeeze-repeats   replace each input sequence of a repeated character
                        that is listed in SET1 with a single occurrence
                        of that character