I have a file abc.fil whose content is in the format yyyymmdd.xyz.doc. I want to cat that file abc.fil such that I get the output as 20150127.xyz.doc. Can any one please help me to resolve this problem.
Sample file:
$ cat abc.fil
o/p: yyyymmdd.xyz.doc
yyyymmdd.mno.dat
abc.txt
Output expected: Instead of yyyymmdd I want the current date to be displayed in the same order.
You could use sed:
sed "s/yyyymmdd/$(date '+%Y%m%d')/g" abc.fil
That replaces the string yyyymmdd with the current date formatted as desired.
Edit:
If yyyymmdd is just the format of the date you want to replace, then use that command (assumes GNU sed):
sed -r "s/[12][0-9]{3}[01][0-9][0-3][0-9]/$(date '+%Y%m%d')/g" abc.fil
The long regular expression pattern means the following: The first digit can be 1 or 2 ([12]), the next three can be everything from 0 to 9 ([0-9]), that's the year. Now the month: the first digit can be 0 or 1 and the second can be everything from 0 to 9 ([01][0-9]). And at last the same with the day.
> today=$(date +%Y%m%d)
> sed "s/yyyymmdd/${today}/g" abc.fil
o/p: 20150127.xyz.doc
20150127.mno.dat
abc.txt
try
cat abc.fil | sed s/yyyymmdd/`date +%Y%m%d`/
$IFS doesn't contain digits (cat is redundant though).
$(...) instead of backticks. Otherwise, it works.