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Unable to convert date into no. using date -d +%s syntax in ksh shell

 
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Top Forums UNIX for Dummies Questions & Answers Unable to convert date into no. using date -d +%s syntax in ksh shell
# 8  
Old 06-06-2012
What is the script in post #7 intended to do? Hard to guess from uncommented faulty code.
Are you trying to find out the age of each file or something?
# 9  
Old 06-07-2012
Hi Methyl,
I have made the function which assigns numeric value to each mon.
then i am comparing the date of each file with todays date and i intend to gunzip all the files which are not of todays date.

comparison code i have removed for now,i m just tryin to print month date and filenames but its givin error related to arrays.
below is code for gunzipping files which is removed.

code :

Code:
v1=`ls -lrt | awk '{print $6,$7,$8,$9}'| egrep -v "gz$" |  egrep "MDCR.*" | wc -l`
v2=`date | awk '{print $2}'`
v3=`date | awk '{print $3}'`
datecalc $v2
v4=$?
i=1
while [ $i -le $v1 ]
do
A[$i]=`ls -lrt | awk '{print $6,$7,$8,$9}'| egrep -v "gz$" |  egrep "MDCR.*" | awk '{print $1}' | head -$i | tail -1`
B[$i]=`ls -lrt | awk '{print $6,$7,$8,$9}'| egrep -v "gz$" |  egrep "MDCR.*" | awk '{print $2}' | head -$i | tail -1`
C[$i]=`ls -lrt | awk '{print $9}'| egrep -v "gz$" |  egrep "MDCR.*" | awk '{print $1}' | head -$i | tail -1`
x=${A[$i]}
datecalc $x
v5=$?
y=${B[$i]}
if [ $v5 -lt $v4 ]
then
gzip ${A[$i]}
elif [ $v5 = $v4 && $y -lt $v3 ]
then
gzip ${A[$i]}
fi
i=`expr $i + 1`
done

Moderator's Comments:
Mod Comment Please use [code]...[/code] tags. Video tutorial on how to use them


---------- Post updated 06-07-12 at 03:53 AM ---------- Previous update was 06-06-12 at 10:22 AM ----------

Hi All,

Any luck with above queries.
Last edited by Scrutinizer; 06-06-2012 at 12:46 PM.. Reason: code tags
# 10  
Old 06-07-2012
The code posted in post #9 does not tie up with the explanation. It appears to be code to gzip files rather than to gunzip files.
Are you actually trying to gzip files which are not dated today and which have not already been zipped?

Need to understand about your directory structure. I notice that you have ls -lar. Does this mean that there are files in subdirectories or would ls -la give exactly the same output?

Are you going to run this script manually or from cron. If it is to be run from cron daily, what time will it run?

I think that the script as posted does not work if any file is dated last year (which changes the format of ls -la). Consider what would happen if the script ran on Jan 1st .

Let's clear up what the script is meant to do, then look at a total rewrite.
# 11  
Old 06-07-2012
Hi Methyl,
sorry for typo ,the code is for gzipping the files.
ls -lart gives same o/p as ls -lrt

absolutely corrct ,i am trying to gzip files older than today which are of a certain pattern.
I will be setting a cron for this which will run at 2300 hrs everyday.
My script comapres the month first with todays date,then the day.
this script will fail on 1st jan as jan is assigned a no. 1 which is less than 12(dec),
so all files for dec 31st will no get gzipped.

also Please find the full code for sun OS 5.8

Code:
 
datecalc()
{
if [ $1 = "Jan" ]
then
return 1;
elif  [ $1 = "Feb" ]
then
return 2;
elif  [ $1 = "Mar" ]
then
return 3;
elif  [ $1 = "Apr" ]
then
return 4;
elif [ $1 = "May" ]
then
return 5;
elif  [ $1 = "Jun" ]
then
return 6;
elif  [ $1 = "Jul" ]
then
return 7;
elif  [ $1 = "Aug" ]
then
return 8;
elif [ $1 = "Sep"
then
return 9;
elif  [ $1 = "Oct" ]
then
return 10;
elif  [ $1 = "Nov" ]
then
return 11;
elif  [ $1 = "Dec" ]
then
return 12;
fi
}
v1=`ls -lrt | awk '{print $6,$7,$8,$9}'| egrep -v "gz$" |  egrep "MDCR.*" | wc -l`
v2=`date | awk '{print $2}'`
v3=`date | awk '{print $3}'`
datecalc $v2
v4=$?
i=1
while [ $i -le $v1 ]
do
A[$i]=`ls -lrt | awk '{print $6,$7,$8,$9}'| egrep -v "gz$" |  egrep "MDCR.*" | awk '{print $1}' | head -$i | tail -1`
B[$i]=`ls -lrt | awk '{print $6,$7,$8,$9}'| egrep -v "gz$" |  egrep "MDCR.*" | awk '{print $2}' | head -$i | tail -1`
C[$i]=`ls -lrt | awk '{print $9}'| egrep -v "gz$" |  egrep "MDCR.*" | awk '{print $1}' | head -$i | tail -1`
x=${A[$i]}
datecalc $x
v5=$?
y=${B[$i]}
if [ $v5 -lt $v4 ]
then
gzip ${A[$i]}
elif [ $v5 = $v4 && $y -lt $v3 ]
then
gzip ${A[$i]}
fi
i=`expr $i + 1`
done

thanks
jcpratap
# 12  
Old 06-07-2012
Total rewrite using find rather than ls .
Create a reference file timestamped at the start of day and then look for files of the right name which are older than the reference file.

Code:
YYYYMMDD="`date +%Y%m%d`"       # Reversed date yyyymmdd
REFERENCE=/tmp/reference.${YYYYMMDD}.$$
touch -t ${YYYYMMDD}0000 ${REFERENCE}   # First thing today
#
cd /CP2_logs/wls21/SAS/MDCR/Applicationlogs/WLS13
find . -type f \( -name 'MDCR.*' -a ! -name '*.gz' -a ! -newer ${REFERENCE} \) -print | while read FILENAME
do
                # Remove echo when tested
                echo gzip "${FILENAME}"
done
rm -f "${REFERENCE}"

# 13  
Old 06-07-2012
Hi Methyl,

THANKS A LOT.

very lucid and effecient code,learned new things from this code.

thanks
jcpratap
 
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