How to replace spaces and slash in string in bash?

Question

Providing a string:

foo='Hello \ World! \ x we are friends here we are' 

Suppose there are tabs mixed with spaces after or before the \ character. I want to replace spaces, tabs, and slashes with just spaces. I tried:

 echo "$foo" | tr "[\s\t]\\\[\s\t]\n\[\s\t]" " " | tr -s " " 

Return:

 Hello World! x we are friend here we are 

And I need the result:

 Hello World! x we are friends here we are 

Some ideas, tips or tricks to do this? Can I get the result that I want to get only in the team?

Answer 1

The following single-line layer gives the desired result:

 echo "$foo" | tr '\n' '\r' | sed 's,\s*\\\s*, ,g' | tr '\r' '\n' Hello World! we are friends here we are 

Explanation:

tr '\n' '\r' removes newlines from input to avoid sed special behavior for newlines.

sed 's,\s*\\\s*, ,g' converts spaces with an embedded \ into one space.

tr '\r' '\n' returns unchanged newline strings.

Answer 2

Try the following:

 #!/bin/bash foo="Hello \ World!" echo $foo | sed 's/[\s*,\\]//g' 

Answer 3

If you want to print the output as given, you just need to:

 foo='Hello \ World!' bar=$(tr -d '\\' <<<"$foo") echo $bar # unquoted! 

 Hello World! 

If you want to compress spaces, as they are stored in a variable, then one of:

 bar=$(tr -d '\\' <<<"$foo" | tr -s '[:space:]' " ") bar=$(perl -0777 -pe 's/\\$//mg; s/\s+/ /g' <<<"$foo") 

The advantage of the perl version is that it only removes backtraces of the line continuation (at the end of the line).

---

Note that when you use double quotes, the shell takes care of the continuation of the line (correct, without spaces after the slash:

 $ foo="Hello \ World" $ echo "$foo" Hello World 

So, at this point it is too late.

If you use single quotation marks, the shell will not interpret line extensions, but

 $ foo='Hello \ World! here we are' $ echo "$foo" Hello \ World! here we are $ echo "$foo" | perl -0777 -pe 's/(\s*\\\s*\n\s*)/ /sg' Hello World! here we are 

Answer 4

 foo='Hello \ World! \ x we are friends here we are' 

If you use double quotes, then the shell will interpret the \ character as a line continuation character. Switching to single quotes preserves a literal backslash.

I added a backslash after World! to check for multiple backslash lines per line.

 sed -r ':s; s/( )? *\\ *$/\1/; Te; N; bs; :e; s/\n *//g' <<< "$foo" 

Output:

 Hello World! x we are friends here we are 

What does it do? In pseudo code, you can read this as:

 while (s/( )? *\\ *$/\1/) { # While there a backslash to remove, remove it... N # ...and concatenate the next line. } s/\n *//g # Remove all the newlines. 

In detail, here is what he does:

• :s is the branch labeled s for "start".
• s/( )? *\\ *$/\1/ s/( )? *\\ *$/\1/ replaces the backslash and surrounding spaces. Does it leave one space, if there is one, capturing ( )? .
• If the previous lookup failed, Te goes to label e .
• N concatenates the next line, including the new line \n .
• bs goes back to the beginning. This means that we can process multiple consecutive lines with backslashes.
• :e is the branch labeled e for "end".
• s/\n *//g removes all additional lines of the newline from step 4. It also removes leading spaces from the next line.

Note that T is an extension of GNU. If you need this to work in a different version of sed, you need to use T This is likely to take an extra label b or two.

Answer 5

You can use the read loop to get the result you want.

 arr=() i=0 while read line; do ((i++)) [ $i -le 3 ] && arr+=($line) if [ $i -eq 3 ]; then echo ${arr[@]} elif [ $i -gt 3 ]; then echo $line fi done <<< "$foo" 

Answer 6

With awk :

 $ echo "$foo" Hello \ World! \ x we are friends here we are 

With the ending new line:

 $ echo "$foo" | awk '{gsub(/[[:space:]]*\\[[:space:]]*/," ",$0)}1' RS= FS='\n' ORS='\n\n' Hello World! x we are friends here we are . 

Without a final new line:

 $ echo "$foo" | awk '{ gsub(/[[:space:]]*\\[[:space:]]*/," ",$0) a[++i] = $0 } END { for(;j<i;) printf "%s%s", a[++j], (ORS = (j < NR) ? "\n\n" : "\n") }' RS= FS='\n' Hello World! x we are friends here we are 

Answer 7

sed is a great tool for simple sub-sections on one line, but just use awk for anything else. This uses GNU awk for multi-char RS (with other awks RS='\0' will work for text files that do not contain NUL characters):

 $ echo "$foo" | awk -v RS='^$' -v ORS= '{gsub(/\s+\\\s+/," ")}1' Hello World! x we are friends here we are 

Answer 8

With basinisms such as extended globbing , parameter expansion , etc., but this is probably just as ugly

 foo='Hello \ World!' shopt -s extglob echo "${foo/+( )\\*( )$'\n'/ }" Hello World! 

Answer 9

As I understand it, you just want to remove trailing spaces followed by a backslash with a new line?

In this case, do a search with the regular expression ( ) *\\\n and replace with \1