how to convert date to timestamp for multiple lines by unix shell

Question

I've already greped a log file for multiline of date string:

2012-10-09 14:05:26  
2012-10-09 14:11:18  
2012-10-09 14:12:03

Now I want to convert them to timestamp, and find which one is bigger than a value
date cmd:

date -d "2012-10-09 14:37:59" +%s

I'm not sure how to do it with multiple lines, awk or other shell cmd.

In the format you show, you don't need conversion for making comparisons: string comparisons are okay. — mouviciel
– mouviciel, Commented Oct 9, 2012 at 8:54

Mats Löfström · Accepted Answer · 2012-10-09 08:58:39Z

3

This is just a simple example, a better solution should include parameter passing, error return testing, and more.

#!/bin/sh

SEARCH=`date -d "2012-10-09 14:11:09" +%s`
while read DATETIME ; do
    THIS=`date -d "$DATETIME" +%s`
    if [ "$SEARCH" -le "$THIS" ] ; then
        echo $DATETIME
    fi
done <<EOD
2012-10-09 14:05:26
2012-10-09 14:11:18
2012-10-09 14:12:03
EOD

edited Oct 9, 2012 at 8:58

answered Oct 9, 2012 at 8:58

Mats Löfström

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Birei · Accepted Answer · 2012-10-09 09:06:03Z

1

Using awk:

awk -v d="2012-10-09 14:07:26" '
    BEGIN { 
        FS = "[-: ]"; 
        split( d, arr, /[-: ]/ );
        date_timestamp = mktime( arr[1] " " arr[2] " " arr[3] " " arr[4] " " arr[5] " " arr[6] );
        if ( date_timestamp == -1 ) {
            printf "%s\n", "Bad format for input date";
            exit 1;
        }
    }
    { 
        if ( mktime( $1 " " $2 " " $3 " " $4 " " $5 " " $6 ) > date_timestamp ) { 
            print $0; 
        } 
    }
' infile

It yields:

2012-10-09 14:11:18  
2012-10-09 14:12:03

answered Oct 9, 2012 at 9:06

Birei

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