how to regex replace before colon? - regex

Question

this is my original string:
NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
I want to only add back slash to all the spaces before ':'
so, this is what I finally want:
NetworkManager/system\ connections/Wired\ 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
I need to do this in bash, so, sed, awk, grep are all ok for me.
I have tried following sed, but none of them work
echo NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1 | sed 's/ .*\(:.*$\)/\\ .*\1/g'
echo NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1 | sed 's/\( \).*\(:.*$\)/\\ \1.*\2/g'
echo NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1 | sed 's/ .*\(:.*$\)/\\ \1/g'
echo NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1 | sed 's/\( \).*\(:.*$\)/\\ \1\2/g'
thanks for answering my question.
I am still quite newbie to stackoverflow, I don't know how to control the format in comment.
so, I just edit my original question
my real story is:
when I do grep or use cscope to search keyword, for example "address1" under /etc folder.
the result would be like:
./NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
if I use vim to open file under cursor, suppose my vim cursor is now at word "NetworkManager",
then vim will understand it as
"./NetworkManager/system"
that's why I want to add "\" before space, so the search result would be more vim friendly:)
I did try to change cscope's source code, but very difficult to fully achieve this. so have to do a post replacement:(

Answer 1

If you only want to do the replacements if there is a : present in the string, you can check if there are at least 2 columns, setting the (output)field separator to a colon.
Data:
cat file michaelvandam#Michaels-MacBook-Pro
NetworkManager/system connections/Wired 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
NetworkManager/system connections/Wired 1.nmconnection 14 address1=10.1.10.71/24,10.1.10.1%
Example in awk:
awk 'BEGIN {FS=OFS=":"}{if(NF>1)gsub(" ","\\ ",$1)}1' file
Output
NetworkManager/system\ connections/Wired\ 1.nmconnection:14 address1=10.1.10.71/24,10.1.10.1
NetworkManager/system connections/Wired 1.nmconnection 14 address1=10.1.10.71/24,10.1.10.1

Answer 2

This could be simply done in awk program, with your shown samples, please try following.
awk 'BEGIN{FS=OFS=":"} {gsub(/ /,"\\\\&",$1)} 1' Input_file
Explanation: Simple explanation would be, setting field separator and output field separator as : for this program. Then in main program using gsub(Global substitution) function of awk. Where substituting space with \ in 1st field only(as per OP's remarks it should be done before :) and printing line then.

Answer 3

An idea for a perl one liner in bash to use \G and \K (similar #CarySwoveland's comment).
perl -pe 's/\G[^ :]*\K /\\ /g' myfile
See this demo at tio.run or a pattern demo at regex101.

Answer 4

This might work for you (GNU sed):
sed -E ':a;s/^([^: ]*) /\1\n/;ta;s/\n/\\ /g' file
Replace spaces before : by newlines then replace newlines by \ 's.
Alternative using the hold space:
sed -E 's/:/\n:/;h;s/ /\\ /g;G;s/\n.*\n//' file
Split the line on the first :.
Amend the front section, remove the middle and append the unadulterated back section.

Answer 5

My answer is ugly and I think RavinderSingh13's answer is THE ONE, but I already took the time to write mine and it works (It's written step by step, but it's a one line command):
I got inspired by HatLess answer:
first get the text before the : with cut (I put the string in a file to make it easy to read, but this works on echo):
cut -d':' -f1 infile
Then replace spaces using sed:
cut -d':' -f1 infile | sed 's/\([a-z]\) /\1\\ /g'
Then echo the output with no new line:
echo -n "$(cut -d':' -f1 infile | sed -e 's/\([a-z]\) /\1\\ /g')"
Add the missing : and what comes after it:
echo -n "$(cut -d':' -f1 infile | sed -e 's/\([a-z]\) /\1\\ /g')" | cat - <(echo -n :) | cat - <(cut -d':' -f2 infile)