Question: I needed to fetch first block in "*.*" string (i.e '\*') in shell script. I found that instead of printing the string it is displaying list of all file in current directory. 

How overcome this issue?

main purpose is change string "asterisk.asterick" (this read from /etc/rsyslog.conf) to "\*.debug", here first field i need keep as it is, only second field needs to change.

```
root@xyz-node:/tmp# ccvv=*.*
root@xyz-node:/tmp# echo ccvv
ccvv
root@xyz-node:/tmp# echo $ccvv
new_script.sh rsyslog.conf rsyslog.confe tmp.8R7bE0tYbg
root@xyz-node:/tmp#
root@xyz-node:/tmp#
root@xyz-node:/tmp# echo $ccvv|cut -d "." -f1
new_script

```

Answer: 
Actually you have to quote variable itself. 
`echo "$ccvv"`
As i understand it...

```
$ echo "*.*"
*.*

$ ccvv="*.*"

$ echo $ccvv
*.*

```

Or?