Script for automatic count the file i have

UNIX Solaris
1

Hi

I have a folder will all the file ftp in everyday. I have to go in every morning to check to see everynight have all the 2000+ file have been all FTP in.

so every morning i will check like this

# ls -l *_20080904*_20080904* |wc

if there is 2714 that mean is correct.

File name: summer_3024190_200809041200_2008090413000.csv

Can someone help me to generate a script so that everyday i just need to check the log file will do?

Thanks in adv

Summer

2

Try this.. If I dont get your question wrong... :o

#!/bin/ksh
cd /
lines=`ls -l *_20080904*_20080904* |wc -l`
if [ ${lines} -eq 2714 ]
then
echo "Good"
exit 1
else
echo "Bad"
fi

3

Hi

erm well 20080904 is a data of yesterday. can the script grep yesterday automatic?

sumemr

4

Do you have the GNU date command? If so you can try:

YESTERDAY=$(date --date yesterday +%Y%m%d)
lines=$(ls *_${YESTERDAY}*_${YESTERDAY}* |wc -l)
5

Sorry, I really don't understand your statement

6

Humm i guess i don have GNU date command.
is there other way?

Summer

7

Perhaps you could say what operating system and shell you are using so we can better answer your question.

8

i'm using solaris 9 and ksh.

9

i'm using solaris 9 & ksh.

10

ksh works for all and I dont foresee any issues with that. Maybe you can be clear enough to elaborate what you want to achieve., at least by providing some sample outputs..? :confused:

11

Try this perhaps:

#!/bin/ksh
cd /some/folder
YESTERDAY=$(perl -e 'use POSIX;print strftime("%Y%m%d\n",localtime(time-86400));')
lines=$(ls *_${YESTERDAY}*_${YESTERDAY}* |wc -l)
if [ "${lines}" -eq 2714 ]
then
    echo "Good"
else
    echo "Bad"
    exit 1
fi
12

Thank all for the help. Annihilannic you script works^^

Thanks thanks..
Summer

13

I had a similar problem where I had to get a value based on yesterday. I came up with the following script. I'm sure it could be shortened a bit but it works for me. Here you go...

#!/usr/bin/ksh
#
#

# Use the built-in time command to get the epoch
# time of yesterday. 86400=1 day in seconds.

YESTER_EPOCH=`/usr/bin/perl -e 'printf"%d\n",(time-86400);'`

# Now convert the epoch time into something more usable.

YESTERDAY_TEMP=`perl -e "print scalar(localtime(${YESTER_EPOCH}))"`

# Now parse out the month, day, and year.

MONTH_TEMP=`print ${YESTERDAY_TEMP} | cut -d" " -f2`

DAY_TEMP=`print ${YESTERDAY_TEMP} | cut -d" " -f3`

YEAR=`print ${YESTERDAY_TEMP} | cut -d" " -f5 | cut -c3-4`

# Convert the month text into a number format

case "${MONTH_TEMP}" in
Jan)
MONTH=01
;;
Feb)
MONTH=02
;;
Mar)
MONTH=03
;;
Apr)
MONTH=04
;;
May)
MONTH=05
;;
Jun)
MONTH=06
;;
Jul)
MONTH=07
;;
Aug)
MONTH=08
;;
Sep)
MONTH=09
;;
Oct)
MONTH=10
;;
Nov)
MONTH=11
;;
Dec)
MONTH=12
;;
esac

# Convert the day to 2 digits if not already

case "${DAY_TEMP}" in
1)
DAY=01
;;
2)
DAY=02
;;
3)
DAY=03
;;
4)
DAY=04
;;
5)
DAY=05
;;
6)
DAY=06
;;
7)
DAY=07
;;
8)
DAY=08
;;
9)
DAY=09
;;
*)
DAY=${DAY1_TEMP}
esac

# From here you can continue with the code that INCREDIBLE posted above.

MYDATE=`ls -l *_${YEAR}${MONTH}${DAY}*_${YEAR}${MONTH}${DAY}* | wc -l`

if [ ${MYDATE} -eq 2714 ]
then
echo "Good"
exit 1
else
echo "Bad"
fi

Hope this is what you're trying to do.

Bruhn

14

Oh Sorry.... I didn't see page 2. Someone already posted the answer.