awk within variable 1

[TSch]

Hello everyone,

could you help me out with a little problem (I'm quite new to perl):

I got the following output:
(Note: I'm using AIX / Linux)

ls -all /datastore/backup

-rw-r--r-- 1 root root 6431369457 Jul 10 07:57 /datastore/backup

What I'd like to do is store only the "Jul 10" in a variable ...

Cutting it out is easy:

ls -all /datastore/backup | awk '{print $6 " " $7}'

But how do I get this stored within a variable ?

Using the following stores the whole string within the variable and NOT only the "Jul 10":

$dat2=`ls -all /datastore/backup | awk '{print $6 " " $7}'`;
print $dat2, "\n";

-rw-r--r-- 1 root root 6431369457 Jul 10 07:57 /datastore/backup

Regards,
Thomas Schmitt

 

[feherke]

Hi

Just to correct your code, escape the dollar signs ( $ ) :

[navy]$dat2[/navy][teal]=[/teal]`ls [teal]-[/teal]all [green][i]/datastore/[/i][/green]backup [teal]|[/teal] awk [green][i]'{print [highlight]\[/highlight]$6 " " [highlight]\[/highlight]$7}'[/i][/green]`[teal];[/teal]

But would be preferable to forget such shell scripting practices when coding in Perl.

In case /datastore/backup is a single file to check, this will print its modification time in local date time format :

[navy]@stat[/navy][teal]=[/teal][b]stat[/b] [green][i]'/datastore/backup'[/i][/green][teal];[/teal]
[b]print[/b] scalar localtime [navy]$stat[/navy][teal][[/teal][purple]9[/purple][teal]];[/teal]

Feherke.
[link feherke.github.com/]

http://feherke.github.com/

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