Replace all control characters in a range of line with awk - bash

Question

I got a file with several lines. Some of these lines contain LFs (0x0A) and CRs (0x0D), which I want to get removed. The point is, that I want to replace them with SPACE them only in a range of characters of every line, eg in a File:
30 30 30 30 30 30 30 30 30 30 **0D 0A** 30 30 0A; 0000000000..00
30 30 30 30 30 30 30 30 **0D 0A** 30 30 30 30 0A; 00000000..0000
I want to remove 0d, 0a from position 0 to 12 in every line of the file.
I got
awk '{l=substr($0,1,12);r=substr($0,13);gsub(/\x00-\1F/," ",l);print l r}' ${f} > ${f}.noLF
but this seems not to work. I guess substr stops at the first 0x0d.
Is there another solution?

Answer 1

awk '/\r$/ && length < 13 {sub(/\r$/,""); printf "%s ", $0; next} {print}' file

Answer 2

Here is something ugly that may work!
Save it as go
#!/bin/bash
while :
do
# Read 13 bytes from stdin, and replace carriage returns and linefeeds with spaces
dd bs=13 count=1 2>/dev/null | tr '\r\n' ' '
# Break out of loop if dd was not successful
[ ${PIPESTATUS[0]} -ne 0 ] && break
# Get rest of conventional line, breaking out of loop if EOF
read rest || break
echo $rest
done
It reads 13 bytes from your file and removes all carriage returns and linefeeds. Then it reads the rest of the conventional line and outputs that.
Use it like this:
chmod +x go
./go < yourfile
Example:
more file
q
wertyuiopqwertyuiop
qwerty
uiopqwertyuiop
./go < file
q wertyuiopqwertyuiop
qwerty uiopqwertyuiop
EDITED TO MATCH FURTHER QUESTIONS
#!/bin/bash
while :
do
# Read 13 bytes from stdin, and replace carriage returns and linefeeds with spaces
first13=$(dd bs=13 count=1 2>/dev/null)
ddexitstatus=$?
if [ echo $first13 | grep -q "^KT" ]; then
echo $first13
else
echo $first13 | tr '\r\n' ' '
fi
# Break out of loop if dd was not successful
[ $ddexitstatus -ne 0 ] && break
# Get rest of conventional line, breaking out of loop if EOF
read rest || break
echo $rest
done