date from filename

Shell Programming and Scripting
1

Hi all

I have the following question:

With this command, I get the latest file in a directory.
lastfile =`ls -1tr | tail -n 1`
echo $lastfile

The output is then:
partner131210.txt (meaning 13th December 2010)

My goal is to get the date into a variable and to obtain a final variable that contains the difference in days between today and the date in the file name.

Thanks for your help.

Best regards,
David

2

To cut date from file name use cut command.

fileDate=`echo partner131210.txt | cut -c8-13`

And this links will help u to find date difference. You should search related forums before posting.
link1 link2 link3

3 
echo partner131210.txt | grep -Eo [0-9][0-9]*

---------- Post updated at 04:10 PM ---------- Previous update was at 04:08 PM ----------

I doubt we can use cut -c8-13 here because we are not sure about the how many characters are there before date string. Though here date starts from 8th char.

4

@R0H0N, Will work as long as date is always in 8th-13th position, Won't work if file name itself has some number like abc123131210.txt OR abcdefgabc131210.txt

 
typeset -i a #Date String from the file name
typeset -i b #Day value from the file name
typeset -i c #Day value from current date
typeset -i d #Difference in days
a=$(ls -1tr | tail -n 1 | sed 's/.*\(......\)\..*/\1/')
b=$(echo $a | cut -c1-2)
c=$(date '+%m')
d=$b-$c
5

May be,

$ lastfile=partner131210.txt
$ expr $lastfile : '.*\(......\).txt'
131210
$