regex - Search and replace in bash using regular expressions

Question

I've seen this example:

hello=ho02123ware38384you443d34o3434ingtod38384day
echo ${hello//[0-9]/}

Which follows this syntax: ${variable//pattern/replacement}

Unfortunately the pattern field doesn't seem to support full regex syntax (if I use . or s, for example, it tries to match the literal characters).

How can I search/replace a string using full regex syntax?

Answer 1

Use sed:

MYVAR=ho02123ware38384you443d34o3434ingtod38384day
echo "$MYVAR" | sed -e 's/[a-zA-Z]/X/g' -e 's/[0-9]/N/g'
# prints XXNNNNNXXXXNNNNNXXXNNNXNNXNNNNXXXXXXNNNNNXXX

Note that the subsequent -e's are processed in order. Also, the g flag for the expression will match all occurrences in the input.

You can also pick your favorite tool using this method, i.e. perl, awk, e.g.:

echo "$MYVAR" | perl -pe 's/[a-zA-Z]/X/g and s/[0-9]/N/g'

This may allow you to do more creative matches... For example, in the snip above, the numeric replacement would not be used unless there was a match on the first expression (due to lazy and evaluation). And of course, you have the full language support of Perl to do your bidding...