Split a string by multiple delimeters in bash

Question

There's a string:

onetwothree.file.001.txt ; threefourfive.file.0.98.txt ; fivefoursix.file.14.txt

I want to split it by . and ; removing the prefix before the filename, so that it looks like this:

file.001.txt ; file.0.98.txt ; file.14.txt

Any ideas?

Answer 1

sed -e 's/[^.]*.//' -e 's/;[^.]*./; /g'

This initially removes the shortest substring up to . from the beginning and then can rely on ; to operate on the resulting string.

Answer 2

Taking the "in bash" literally, you could do something like this.

• split the string into an array, delimited by semicolons

• remove the prefix elementwise and store the result in a string, delimited by the first character of IFS

• globally add back whitespace behind the delimiters

NOTE: you may wish to save the current IFS so you can restore it afterwards.

IFS=";"
read -a arr <<< "onetwothree.file.001.txt ; threefourfive.file.0.98.txt ; fivefoursix.file.14.txt"
printf -v str "${arr[*]#*.}"
printf "%s\n" "${str//;/; }"

ginving

file.001.txt ; file.0.98.txt ; file.14.txt

Answer 3

Or, with sed...

s="onetwothree.file.001.txt ; threefourfive.file.0.98.txt ; fivefoursix.file.14.txt"
sed -E "s/(^|; )[^\.]+\./\1/g" <<<$s

Walkthrough

(^|; )[^\.]+\.

Find any sub-element that either starts either at the beginning of the line ^ or | with ; (semicolon and a space) and which is followed by [^\.]+\. i.e. a continuous series which does not contain a literal . but which does end with a literal .

Then replace all of that with \1 which is the capture group (^|; )

Output

file.001.txt ; file.0.98.txt ; file.14.txt