UNIX for Dummies Questions & Answers

I am using KSH.

Code:

 
d=date | nawk '{ print $2,$3-1}'
ls -lrt SystemOut* | nawk '/d/{print $NF}'

I am trying to list the files of previous day but it doesn't seem to be working and i am also not getting an error.

result is blank.Need help.

Code:

d=date | nawk '{ print $2,$3-1}' 
ls -lrt SystemOut* | nawk '$0 ~ d {print $NF}' d="${d}"

What will happen on the 1-st of a month?
What will happen on the 1-st of March of a leap/non-leap year?

Hi Vgresh,

Could you please explain what does d="${d}" do?

Were you asking a question to me in the above post ?

---------- Post updated at 01:13 PM ---------- Previous update was at 11:17 AM ----------

Code:

d=`date | nawk '{ print $2,$3-1}'`
ls -lrt SystemOut* | nawk '$0 ~ d {print $NF}' d="${d}"

yes, I was asking you..........

Code:

 d="${d}"

This assigns the value of variable $d defined in script as the previous day to variable d in awk.

Code:

 $0 ~ d

The above is a containment test which will be true if the value of d matches anywhere on the line returned from ls command.

To match only those files that were created/modified based in the value of $d, you should use field $7 (the day), not $0.

Code:

 ls -lrt SystemOut* | nawk '$7 ~ d {print $NF}' d="${d}"

The above exectutes fine on Solaris.

Note that vgersh99's questions still apply.

If you don't need awk, try this:

Code:

find . -daystart -mtime -2 ! -mtime -1 -ls

Will work in leap years and on month's first...

vgersh99,
Nothing will happen with the code you supplied.

If you change it to:

Code:

d=$(date | nawk '{ print $2,$3-1}')
ls -lrt SystemOut* | nawk -v d="${d}" '$0 ~ d {print $NF}'

It will find the files you want if you run it on days 11-31 in any month.
The first line calls nawk to convert the output of the date command (which today would be something like "Sun Jul 29 10:07:09 PDT 2012") to "Jul 28".

But, it will also list any other lines that contain "Jul 28" in the file name instead of in the date field. It won't work on the day 1-10 of any month because the awk command:

Code:

print $2,$3-1

puts one space between the two values being printed, but the ls -l output has two spaces between the abbreviated month and the day for days 1 through 9 each month (which will keep the match of $0 to the awk variable d to fail).

Both of these problems can be fixed by keeping the abbreviated month name and the day of the month separate and matching them agains the 6th and 7th fields, respectively, in the output of ls instead of matching the combined fields agains the entire line.

But, it still won't work on the first of any month because the first command on the first of any month will set d to <MonthAbbreviation><space>0 and you'll never find ls -l output with a 0 as the day. So, using this method, you need to make your first awk command smart enough to convert the things like "Sun Jul 1 10:07:09 PDT 2012" to "Jun 30" instead of "Jul 0". If you have understood this much, I assume you now understand why leap years add to this complexity when calculating the previous day for March 1st.

The find -daystart primary suggested by RudiC with -mtime 1 would take care of this on some implementations, but I believe you said on another thread that you were running on OS/X and awk on OS/X doesn't provide the daystart primary.

On OS/X, the following should do what you want:

Code:

#!/bin/ksh
touch datestamp$$
touch -A -240000 datestamp$$
set -- $(ls -l datestamp$$)
rm datestamp$$
ls -lrt SystemOut*|awk -v m=$6 -v d=$7 '$6 ~ m && $7 ~ d {print $NF}'