Bash script variables comparision

[linux777]

Hi people!
I would like to start by saying that I'm new to bash scripting! I'm trying to write a script that could list all files that have been modified today an that are bigger then 400k recursively from the location specified in the argument.

The problem is that the script does not return any results !!
so when i replace the $date with "2010-01-30" the script returns some results !!

I'm quite sure that the problem comes from the comparison of ($6 == $dat)

#!/bin/bash


if [ -z $1 ]
then
echo "Enter a path: "

else [ -n $1 ]


dat=`date +%F`

path1=$1


file="$(ls $path1 -lR | awk '{if(($5>400)&&($6 == $dat))print $5 "k\t" $6 "\t" $8 "\n\v\r"}')"
echo $file

fi

Please tell me what i'm i doing wrong

[scathefire]

i'll tell you right now your date format is wrong.

$6 (Jan 17:25) will not equal date +%F (2010-01-30). there are a few ways around it, what i would do is add --full-time to my ls command. your ls output will look like this:

[user1@dumbbot ~]$ ls -lR --full-time /home/user1
/home/user1:
total 21416
-rw-r--r-- 1 user1 wheel 21887319 2010-01-30 17:25:34.000000000 -0600 cpinfo-00034076.txt
-rwx------ 1 user1 wheel      196 2010-01-30 18:10:12.000000000 -0600 new-script.sh

[linux777]

Thanks for the quick answer!

when i run the ls command in my terminal, i get the following output

kalle@ubuntuFS:~/uppgift7$ ls -l
total 196
drwxr-xr-x 2 kalle kalle  4096 2010-01-29 00:36 awk1
-rwxr-xr-x 1 kalle kalle   562 2010-01-27 23:00 exp1.sh
-rwxr-xr-x 1 kalle kalle   415 2010-01-27 23:38 exp2.sh

so the $6 in this case is 2010-01-27.

The output from ls -lR --full-time

kalle@ubuntuFS:~/uppgift7$ ls -lR --full-time
.:
total 196
drwxr-xr-x 2 kalle kalle  4096 2010-01-29 00:36:48.000000000 +0100 awk1
-rwxr-xr-x 1 kalle kalle   562 2010-01-27 23:00:40.000000000 +0100 exp1.sh

so i'v changed the script as you recommended

#!/bin/bash
if [ -z $1 ]
then
echo "Enter a path: "
else [ -n $1 ]
dat=`date +%F`
path1=$1

file="$(ls $path1 -lR --full-time | awk '{if(($5>400)&&($6 == $dat))print $5 "k\t" $6 "\t" $8 "\n\v\r"}')"
echo $file
fi

This is what i got

kalle@ubuntuFS:~/uppgift7$ ./uppgift15.sh /home/kalle/uppgift7/

kalle@ubuntuFS:~/uppgift7$

and if i change the $dat to "2010-01-30" in the script, then the output becomes like this (like i'm expecting it to be!)

kalle@ubuntuFS:~/uppgift7$ ./uppgift15.sh /home/kalle/uppgift7/
4096k 2010-01-30 +0100 
 448k 2010-01-30 +0100 
 513k 2010-01-30 +0100 
 513k 2010-01-30 +0100 

kalle@ubuntuFS:~/uppgift7$

Any ideas? Help appreciated!

[scathefire]

sorry on i was testing your script on a centos box, and the timestamp format is different.

try using eq instead of ==

[linux777]

Well the eq instead of == gave some results but the output is still wrong

kalle@ubuntuFS:~/uppgift7$ ./uppgift15.sh /home/kalle/uppgift7/
4096k 2010-01-29 +0100 
 562k 2010-01-27 +0100 
 415k 2010-01-27 +0100 
 4096k 2010-01-30 +0100 
 4096k 2010-01-29 +0100 
 556k 2010-01-29 +0100 
 12243k 2010-01-29 +0100 
 11777k 2010-01-29 +0100 
 448k 2010-01-30 +0100 
 580k 2010-01-28 +0100 
 513k 2010-01-30 +0100 
 513k 2010-01-30 +0100 
 4096k 2010-01-29 +0100 

kalle@ubuntuFS:~/uppgift7$

Thanks for trying!

[scathefire]

why do you have to do it that way? why not use a command like find?

find $1 -type f -daystart -size +40k

where $1 is your directory, otherwise it will run the command in the pwd

[nmset]

You may consider :

find "$1" -type f -newermt 00:00 -size +400k -printf "%p %T+ %u %g %s\n"

man find

EDIT:
If you use daystart, you must provide a further test on a time value, else find will list everything

find $1 -type f -daystart -size +400k -mtime 0 #-printf "%p %T+ %u %g %s\n"

[linux777]

Thanks for the alternative solutions! worked just fine. But still have to do this script with the AWK command to fulfill the conditions of the lab!

When i echo the $dat, it returns today's date 2010-01-31!
The print $6 from the output of the AWK command returns a date in the same format "2010-01-*". some days do match the $6 output "2010-01-31". In this case the comparison ($6 == $dat) should return true! !

so what is the problem? I'm very confused!

When i replace the $dat with the string "2010-01-30" ($6 == "2010-01-30"), the script return what i'm looking for.

#!/bin/bash

if [ -z $1 ]
then
echo "Enter a path: "

else [ -n $1 ]


dat=`date +%F`
echo $dat
path1=$1
echo $path1

file="$(ls $path1 -lR --full-time | awk '{if(($5>400)&&($6 == "2010-01-30"))pri$
echo $file

Thanks again for the help!