Analyzing the CLI output with AWK

[brunocam]

Hi, I'm trying to creat a script to organize a bunch of files in directories of years and month, and for this I will use awk and shell scripts, but I would like to know if I can use the awk's output to feed the functions of my script, becouse I had used awk before but when I typed for exemple (ls -l|awk '{print $6"-"$8}') to select only the important columns I send all to a text file ( >> log.txt) and after this the script had started to analyse the lines, but I would like to do that without create this file text. Is possible to do this with AWK.

 

[Annihilannic]

You can use awk to print the commands, and pipe them to another shell:

ls -l | awk '{print "mv "$9" "$6"-"$8"/"$9}' | sh

Or you can run the move commands directly from your awk script using the system() function, which is a bit slower:

ls -l | awk '{system("mv "$9" "$6"-"$8"/"$9)}'

Of course this assumes that the destination directories already exist. It also assumes that no filenames contain spaces.

Remember also that field 8 in ls -l output may contain either a month or a time, depending on how old the file is.

Annihilannic.

 

[brunocam]

cool suggestion, but I still have one doubt...

look, in every thing that I had read always the teach how to take parameters from a text file like this:

awk '$5 == 57 { print $8}' /bruno/log.txt

but I have to organize a directory with thousands and thousandos of files, and if I have to first create a text file with the (ls -l) in this directory for just latter start to analyse this will take a long long time...

So, I must to know is how to take this output directly from the shell prompt. In the command above, how could I replace the 'log.txt' to samply the output of the shell prompt:
# ls -l /bruno

This is the bigger question for me.

 

[Annihilannic]

ls -l /bruno | awk '$5 == 57 { print $8}'

Annihilannic.

 

[brunocam]

Simple and smart...

the output from this command is:

2011-07-24-bruno.sh
2011-07-29-flex.awk
2011-07-29-flex.sh
2011-07-24-staff
2011-07-24-teste2
2011-07-24-teste.awk

May I put this result into a variable?
I had try to do like this:
____________________________________________________
#! /bin/sh

VAR = $(ls -l /bruno | awk '$5 == 57 { print $8}')
----------------------------------------------------
But it doesn't work

 

[Annihilannic]

Include your code between [ignore]

 ...

[/ignore] tags to format it nicely in your posts.

What operating system is it exactly? On many OS /bin/sh is fairly primitive and doesn't understand $( ... ) syntax, in which case you may need to use backquotes instead: ` ... `

Also remove the spaces around the '=' sign.

Annihilannic.

 

[brunocam]

I'm using CentOS

I had try to use backquotes too but it doesn't work...

If I could to put the output line in a variable, it would be much better... and fast...

 

[Annihilannic]

In that case $( ... ) is fine because /bin/sh actually runs bash.

Did you take out the spaces like I suggested, i.e.:

#!/bin/sh

VAR=$(ls -l /bruno | awk '$5 == 57 { print $8}') 

echo "VAR is $VAR"

Annihilannic.