Replace string in unix from 10th column onwards

Linux and Unix-Like UNIX and Linux Applications
1

Hi All,

I need to replace the last 19 bytes of the following string

My_Org_Testing_20090102_231124.txt (Text_Date_Time.txt).

I would like to derive the current time using "date +%Y%m%d_%H%M%S.txt" and replace the last 19 bytes of the above string

I would appreciate if someone could help me on this.

Thanks in advance

Regards,
Rohan

2

Something like this.

var=My_Org_Testing_20090102_231124.txt 
echo $var| sed 's/My_Org_Testing_\(.*\)/My_Org_Testing_'$(date +%Y%m%d_%H%M%S.txt)'/
3 
$ s="My_Org_Testing_20090102_231124.txt"
$ echo ${s%???????????????????}$(date +%Y%m%d_%H%M%S.txt)
4 
 $ echo "My_Org_Testing_20090102_231124.txt"  | sed "s/.\{19\}$/$(date +%Y%m%d_%H%M%S.txt)/"