Could you please try following code and let us know if this helps.
Thanks,
R. Singh
Thank you dear Singh.
but your command doesn't work.
Quote:
Originally Posted by Don Cragun
First, never tell us: "i use your command and make this script but it wouldn't work"! Show us the exact output from running the script that explains to us exactly what didn't work. If you don't show us how it isn't working, it makes it a huge guessing game for all of us (especially those of us who are using a different operating system and shell than you're using).
Second, there is a HUGE difference between running a program and feeding its output to awk as RudiC suggested with:
and running a program, saving its output in a variable, executing a fixed string, and feeding the results of the command named by that string to awk as your code is doing:
Why are you storing the output from ls in a variable.
Third: You have an extraneous single quote before the -F option to awk.
And, fourth: I don't see how this script is going to work if you use ls -l instead of ls without the -l option. (With ls -l output, $1 in awk is never going to match Full nor level.)
Please try running your script more like RudiC suggested:
dear Don . i use the ssh to get the directory info. so i had to run and put it in a variable.
i use your shell command and it work but not the way i want it.
result:
i want just return a value. your shell script return a value for each directory name.
let me put it this way:
i want check the directory for backup status. valid backup lifetime is less than two days. now
if the script return the "10" i found out i have a full backup which is valid.
if the script return the "5" i found out i have a level backup which is valid.
in other wise i dont have a valid backup file.
Quote:
Originally Posted by RudiC
You assigned the ls result to the "output" variable. To get at its contents, the shell needs to expand it: $output, to keep the shell from swallowing redundant spaces etc, quote it: "$output". To pipe it into sth. else, echo it: echo "$output".
Why don't you pipe ls's results immediately, as I alluded in my post?
thank you dear RudiC.
i use the ssh to get the directory info. so i had to run and put it in a variable.
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