[regex] Search and replace in bash using regular expressions

These examples also work in bash no need to use sed:

#!/bin/bash
MYVAR=ho02123ware38384you443d34o3434ingtod38384day
MYVAR=${MYVAR//[a-zA-Z]/X} 
echo ${MYVAR//[0-9]/N}

you can also use the character class bracket expressions

#!/bin/bash
MYVAR=ho02123ware38384you443d34o3434ingtod38384day
MYVAR=${MYVAR//[[:alpha:]]/X} 
echo ${MYVAR//[[:digit:]]/N}

output

XXNNNNNXXXXNNNNNXXXNNNXNNXNNNNXXXXXXNNNNNXXX

What @Lanaru wanted to know however, if I understand the question correctly, is why the "full" or PCRE extensions \s\S\w\W\d\D etc don't work as supported in php ruby python etc. These extensions are from Perl-compatible regular expressions (PCRE) and may not be compatible with other forms of shell based regular expressions.

These don't work:

#!/bin/bash
hello=ho02123ware38384you443d34o3434ingtod38384day
echo ${hello//\d/}


#!/bin/bash
hello=ho02123ware38384you443d34o3434ingtod38384day
echo $hello | sed 's/\d//g'

output with all literal "d" characters removed

ho02123ware38384you44334o3434ingto38384ay

but the following does work as expected

#!/bin/bash
hello=ho02123ware38384you443d34o3434ingtod38384day
echo $hello | perl -pe 's/\d//g'

output

howareyoudoingtodday

Hope that clarifies things a bit more but if you are not confused yet why don't you try this on Mac OS X which has the REG_ENHANCED flag enabled:

#!/bin/bash
MYVAR=ho02123ware38384you443d34o3434ingtod38384day;
echo $MYVAR | grep -o -E '\d'

On most flavours of *nix you will only see the following output:

d
d
d

nJoy!