Re: script command question
I'm trying to improve my shell programming skills anyway, so I made a
first pass at a program that will strip backspaces from an input file. I
stopped short of making it robust enough to handle edits that cross
a newline; for example, the following backspaces will be left in the
Here comes the end of line^H^H^H^H
Here is the output from a sample input file, and the script is included
at the bottom. I apologize for poor line wrapping; I wasn't careful to
keep the line length less than 75 when I wrote the script. Please
comment and offer suggestions for improvement.
tesla:~/bin$ cat -v test.in
Now is the tame^H^H^H^Htime for all good men
to come to the help^H^H^H^Haid of their ct^Hountry.
tesla:~/bin$ cat test.in | rmbs > test.out
tesla:~/bin$ cat -v test.out
Now is the time for all good men
to come to the aid of their country.
# rmbs: Remove backspaces from standard input, replacing the old characters
# with the new, and write to standard output. A BLATANT SHORTCOMING:
# this script will not replace edits that cross a newline
# this might be a bash-ism, if it doesn't work try: bs=`echo -n a | tr a '\010'`
# This sed string matches: zero or more non-backspace characters followed
# by one or more backspace characters, followed by zero or more characters,
# and replaces the entire line with the backspace characters. Basically,
# it outputs the first string of backspace characters to be counted with wc.
while read line
# iteratively find strings of backspaces, because sed will only find
# the first one on a line
while echo "$line" | grep "$bs" > /dev/null
# count the backspaces in the first string of backspaces
n=`echo -n "$line" | sed $find | wc -c`
n=`expr $n` # remove leading spaces
# This sed string replaces the n characters before
# the n backspaces with the n characters following,
# where n is the number of backspace characters
# matched in the previous sed call
# NOTE: no replace if fewer than n characters before
# or after the backspace string
line=`echo "$line" | sed $replace`
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