Remove spaces between patterns - bash

Question

I have a log file where data is separated by spaces. Unfortunately one of the datafields contains spaces as well. I would like to replace those spaces with "%20". It looks like this:
2012-11-02 23:48:36 INFO 10.2.3.23 something strange name.doc 3.0.0 view1 orientation_right
the expected result is
2012-11-02 23:48:36 INFO 10.2.3.23 something%20strange%20name.doc 3.0.0 view1 orientation_right
unpredictable that how many spaces we have between the IP address and ".doc". So I would like to change them between these two patterns using pure bash if possible.
thanks for the help

Answer 1

$ cat file
2012-11-02 23:48:36 INFO 10.2.3.23 something strange name.doc 3.0.0 view1 orientation_right
Using Perl:
$ perl -lne 'if (/(.*([0-9]{1,3}\.){3}[0-9]{1,3} )(.*)(.doc.*)/){($a,$b,$c)=($1,$3,$4);$b=~s/ /%20/g;print $a.$b.$c;}' file
2012-11-02 23:48:36 INFO 10.2.3.23 something%20strange%20name.doc 3.0.0 view1 orientation_right

Answer 2

This might work for you (GNU sed):
sed 's/\S*\s/&\n/4;s/\(\s\S*\)\{3\}$/\n&/;h;s/ /%20/g;H;g;s/\(\n.*\n\)\(.*\)\n.*\n\(.*\)\n.*/\3\2/' file
This splits the line into three, copies the line, replaces space's with %20's in one of the copies and reassembles the line discarding the unwanted pieces.
EDIT:
With reference to the comment below, the above solution can be ameliorated to:
sed -r 's/\S*\s/&\n/4;s/.*\.doc/&\n/;h;s/ /%20/g;H;g;s/(\n.*\n)(.*)\n.*\n(.*)\n.*/\3\2/' file

Answer 3

Untested as of yet, but in Bash 4 it's possible to do this
if [[ $line =~ (.*([0-9]+\.){3}[0-9]+ +)([^ ].*\.doc)(.*) ]]; then
nospace=${BASH_REMATCH[3]// /%20}
printf "%s%s%s\n" ${BASH_REMATCH[1]} ${nospace} ${BASH_REMATCH[4]}
fi

Answer 4

Here's one way with GNU sed:
echo "2012-11-02 23:48:36 INFO 10.2.3.23 something strange name.doc 3.0.0 view1 orientation_right" |
sed -r 's/(([0-9]+\.){3}[0-9]+\s+)(.*\.doc)/\1\n\3\n/; h; s/[^\n]+\n([^\n]+)\n.*$/\1/; s/\s/%20/g; G; s/([^\n]+)\n([^\n]+)\n([^\n]+)\n(.*)$/\2\1\4/'
Output:
2012-11-02 23:48:36 INFO 10.2.3.23 something%20strange%20name.doc 3.0.0 view1 orientation_right
Explanation
s/(([0-9]+\.){3}[0-9]+\s+)(.*\.doc)/\1\n\3\n/ # Separate the interesting bit on its own line
h # Store the rest in HS for later
s/[^\n]+\n([^\n]+)\n.*$/\1/ # Isolate the interesting bit
s/\s/%20/g # Do the replacement
G # Fetched stored bits back
s/([^\n]+)\n([^\n]+)\n([^\n]+)\n(.*)$/\2\1\4/ # Reorganize into the correct order

Answer 5

Just bash. Assuming 4 fields appear before the space separated string and 3 fields appear after:
reformat_line() {
local sep i new=""
for ((i=1; i<=$#; i++)); do
if (( i==1 )); then
sep=""
elif (( (1<i && i<=5) || ($#-3<i && i<=$#) )); then
sep=" "
else
sep="%20"
fi
new+="$sep${!i}"
done
echo "$new"
}
while IFS= read -r line; do
reformat_line $line # unquoted variable here
done < filename
outputs
2012-11-02 23:48:36 INFO 10.2.3.23 something%20strange%20name.doc 3.0.0 view1 orientation_right

Answer 6

A variation on Thor's answers, but using 3 processes (4 with the cat bellow but you can get rid of it by putting your_file as the last argument of the 1st sed):
cat your_file |
sed -r -e 's/ (([0-9]+\.){3}[0-9]+) +(.*\.doc) / \1\n\3\n/' |
sed -e '2~3s/ /%20/g' |
paste -s -d " \n"
As Thor explained:
The 1st sed (s/ (([0-9]+\.){3}[0-9]+) +(.*\.doc) / \1\n\3\n/) separates the interesting bit on its own line.
And then:
The 2nd sed replaces all spaces by %20 on the 2nd line and every 3 lines.
Finally, paste it back together.
It must be noted that the 2~3 part is a GNU sed extension. If you do not have GNU sed, you can do:
cat your_file |
sed -r -e 's/ (([0-9]+\.){3}[0-9]+) +(.*\.doc) / \1\n\3\n/' |
sed -e 'N;P;s/.*\n//;s/ /%20/g;N' |
paste -s -d " \n"