UNIX for Dummies Questions & Answers

Hi,
I do have a file name as follows:

12345.txt


Now in my script I need to have a varibale which will hold the file name alone wihtout any extensions.

VAL=12345


How can this be done ?? The file name size is not fixed.

Code:

x=12345.txt
val=$( echo ${x%.*} )
echo $val

Have You tried basename?

lakris@Lakrix:~/Projects/scripts$ basename 12345.txt
12345.txt
lakris@Lakrix:~/Projects/scripts$ basename 12345.txt .txt
12345
lakris@Lakrix:~/Projects/scripts$ basename /tmp/12345.txt .txt
12345
lakris@Lakrix:~/Projects/scripts$ man basename
Reformatting basename(1), please wait...
lakris@Lakrix:~/Projects/scripts$ dirname /tmp/12345.txt
/tmp
lakris@Lakrix:~/Projects/scripts$

/Lakris

This should be sufficient:

Code:

x=12345.txt val="${x%.*}"

hi radoulov ,

Thank u for the solution. But i need one more help. You have assigned
x=12345.txt

Say I do have some ten files in folder. I need to get the names of those files (without the extensions) . The print the maximum of that.

Say I do have follwoing files in a folder:

111.txt
222.txt
333.txt

I need a script whcih will give an output as :
333

Kindly give a soultion for this .. please

The implementation depends on the shell you're using.

With zsh:

Code:

zsh-4.3.4% ls
1111.txt  111.txt  222.txt  333.txt
zsh-4.3.4% set -- *(:r)
zsh-4.3.4% print ${${(n)@}[-1]}
1111

With bash/ksh93 and sort:

Code:

(IFS=$'\n';set -- $(printf "%s\n" *|sort -rn);printf "%s\n" "${1%.*}")

Otherwise:

Code:

(IFS='
';set -- $(printf "%s\n" *|sort -rn);printf "%s\n" "${1%.*}")

(assuming no embedded new lines in the filenames)
(IFS is set to new line)

Code:

ls | sort -rn | read a; echo ${a%.*}

I think that's the shortest solution (in bytes).