Searching logs in UNIX between To and FROM date in shell script

Question

I have a script that searches logs directory using grep command, below is the command.

export CMD="grep -ie '.*${RUN_DATE}.*${MESSAGE_TO_SEARCH}' ${LOG_FILE_PATH}/${LOG_FILE} | awk -F 'addInfo' '{print \$2;}' | sed '/^$/d'"

Now RUN_DATE is any date on which the script is run.

MESSAGE_TO_SEARCH is any message that we are searching in logs file.

It is declared before starting the script.

I want to modify the script that it searches between TO and FROM time, for eg:

FROM_TIME=2016-08-28 13:30:00 and

TO_TIME=2016-08-29 17:00:00

So the script should give logs that lies within this time only.

I tried the following by using passing dates as regex, but it didn't work.

export CMD="grep -ie '.*[${FROM_DATE}-${TO_DATE}].*${MESSAGE_TO_SEARCH}' ${LOG_FILE_PATH}/${LOG_FILE} | awk -F 'addInfo' '{print \$2;}' | sed '/^$/d'"

Please suggest correct method. Thanks

EDIT1: Please tell me a way i can do this, if regex is not appropriate for it.

Don't store command in a variable. You would need to use eval to use it. — anishsane
– anishsane, Commented Aug 30, 2016 at 7:46
Regex isn't the best tool to match date range. You'd have better luck parsing the log file names with a script. Here is an example showing how complicated a regex can become for dates — Aserre
– Aserre, Commented Aug 30, 2016 at 9:08

diov · Accepted Answer · 2016-08-30 12:29:44Z

1

If you just want to select the lines from a log file, there is another way with some simple commands but more readable.

1.Get the line numbers of the first and the last line.

# SN for start line number, EN for end line number.

export SN=`cat -n LOG_FILE | grep START_TIME | head -1 | awk '{print $1}'`
export EN=`cat -n LOG_FILE | grep END_TIME | tail -1 | awk '{print $1}'`

2.Get the lines you want.

head -$EN LOG_FILE | tail -$(($EN-$SN+1))

Once we know the line number we need, we can use head and tail to cut off the part we don't need. ^_^

answered Aug 30, 2016 at 12:29

diov

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