replace string to asterisk bash

Question

I am trying to get from user a path as an input.
The user will enter a specific path for specific application:

script.sh /var/log/dbhome_1/md5

I've wanted to convert the number of directory (in that case - 1) to * (asterisk). later on, the script will do some logic on this path.

When i'm trying sed on the input, i'm stuck with the number -

echo "/var/log/dbhome_1/md5" | sed "s/dbhome_*/dbhome_\*/g"

and the input will be -

/var/log/dbhome_*1/md5

I know that i have some problems with the asterisk wildcard and as a char... maybe regex will help here?

Answer 1

Code for GNU sed :

sed "s#1/#\*/#"

.

$echo "/var/log/dbhome_1/md5" | sed "s#1/#\*/#"
"/var/log/dbhome_*/md5"

---

Or more general:

sed "s#[0-9]\+/#\*/#"

.

$echo "/var/log/dbhome_1234567890/md5" | sed "s#[0-9]\+/#\*/#"
"/var/log/dbhome_*/md5"

Answer 2

use this instead:

echo "/var/log/dbhome_1/md5" | sed "s/dbhome_[0-9]\+/dbhome_\*/g"

[0-9] is a character class that contains all digits

Thus [0-9]\\+ matches one or more digits

Answer 3

If your script is in bash (which I assume when I see the tag, but I also doubt it when I see its name script.sh which seems to have the wrong extension for a bash script), you might as well use pure bash stuff: /var/log/dbhome_1/md5 will very likely be in positional parameter $1 , and what you want will be achieved by:

echo "${1//dbhome_+([[:digit:]])/dbhome_*}"

If this seems to fail, it's probably because your extglob shell optional behavior is turned off. In this case, just turn it on with

shopt -s extglob

Demo:

$ shopt -s extglob
$ a=/var/log/dbhome_1234567/md5
$ echo "${a//dbhome_+([[:digit:]])/dbhome_*}"
/var/log/dbhome_*/md5
$ 

Done!