Bigger the log file, it will be difficult for anyone to search during analysis, hence wanted to shorten log file based on the Day, Month, and most importantly time(Hour).
I have written script based on the hourly basics, which will retrieve your logs for requested day of the month.
I have tried on Redhat/CentOS.
Script can be found here, Download
#!/bin/bash
LOGFILE="/var/log/messages"
echo "Enter the time stamp to search in log files"
read -p "Day: " DAY
read -p "Month[Eg aug..etc]: " MONTH
read -p "Hour[Eg 02, 10..etc]: " HOUR
echo;
echo -e "\e[00;31mLogs which occured in mentioned timestamp: $DAY"-"$MONTH"-"$HOUR":00" \e[00m"
echo;
if [ $DAY -lt 9 ]
then
BLANK=" "
cat $LOGFILE | grep "$HOUR:[0-5][0-9]" | grep -i -n "$MONTH $BLANK$DAY"
else
cat $LOGFILE | grep "$HOUR:[0-5][0-9]" | grep -i -n "$MONTH $DAY"
fi
LOGFILE="/var/log/messages"
echo "Enter the time stamp to search in log files"
read -p "Day: " DAY
read -p "Month[Eg aug..etc]: " MONTH
read -p "Hour[Eg 02, 10..etc]: " HOUR
echo;
echo -e "\e[00;31mLogs which occured in mentioned timestamp: $DAY"-"$MONTH"-"$HOUR":00" \e[00m"
echo;
if [ $DAY -lt 9 ]
then
BLANK=" "
cat $LOGFILE | grep "$HOUR:[0-5][0-9]" | grep -i -n "$MONTH $BLANK$DAY"
else
cat $LOGFILE | grep "$HOUR:[0-5][0-9]" | grep -i -n "$MONTH $DAY"
fi
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