removing last character of every word in files - regex

Question

I have multiple files with just one line of simple text. I want to remove last character of every word in each file. Every file has different length of text.
The closest I got is to edit one file:
awk '{ print substr($1, 1, length($1)-1); print substr($2, 1, length($2)-1); }' file.txt
But I can not figure out, how to make this general, for files with different words count.

Answer 1

awk '{for(x=1;x<=NF;x++)sub(/.$/,"",$x)}7' file
this should do the removal.
If it was tested ok, and you want to overwrite your file, you can do:
awk '{for(x=1;x<=NF;x++)sub(/.$/,"",$x)}7' file > tmp && mv tmp file
Example:
kent$ awk '{for(x=1;x<=NF;x++)sub(/.$/,"",$x)}7' <<<"foo bar foobar"
fo ba fooba

Answer 2

Use awk to loop till max fields in each row upto NF, and apply the substr function.
awk '{for (i=1; i<=NF; i++) {printf "%s ", substr($i, 1, length($i)-1)}}END{printf "\n"}' file
For a sample input file
ABCD ABC BC
The awk logic produces an output
ABC AB B
Another way by changing the record-separator to NULL and just using print:-
awk 'BEGIN{ORS="";}{for (i=1; i<=NF; i++) {print substr($i, 1, length($i)-1); print " "}}END{print "\n"}' file

Answer 3

I would go for a Bash approach:
Since ${var%?} removes the last character of a variable:
$ var="hello"
$ echo "${var%?}"
hell
And you can use the same approach on arrays:
$ arr=("hello" "how" "are" "you")
$ printf "%s\n" "${arr[#]%?}"
hell
ho
ar
yo
What about going through the files, read their only line (you said the files just consist in one line) into an array and use the abovementioned tool to remove the last character of each word:
for file in dir/*; do
read -r -a myline < "$file"
printf "%s " "${myline[#]%?}"
done

Answer 4

Sed version, assuming word are only composed of letter (if not, just adapt the class [[:alpha:]] to reflect your need) and separate by space and puctuation
sed 's/$/ /;s/[[:alpha:]]\([[:blank:][:punct:]]\)/\1/g;s/ $//' YourFile
awk (gawk for regex boundaries in fact)
gawk '{gsub(/.\>/, "");print}' YourFile
#or optimized by #kent ;-) thks for the tips
gawk '4+gsub(/.\>/, "")' YourFile

Answer 5

$ cat foo
word1
word2 word3
$ sed 's/\([^ ]*\)[^ ]\( \|$\)/\1\2/g' foo
word
word word
A word is any string of characters excluding space (=[^ ]).
EDIT: If you want to enforce POSIX (--posix), you can use:
$ sed --posix 's/\([^ ]*\)[^ ]\([ ]\{,1\}\)/\1\2/g' foo
word
word word
This \( \|$\) changes to \([ ]\{,1\}\), ie there is an optional space in the end.