Calculating using date


 
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Top Forums Shell Programming and Scripting Calculating using date
# 8  
Old 01-28-2011
Thanks, I just figured it out nearly same way.
Code:
#!/bin/bash
echo -n  "status.value "

t1=$(wget -q -O - "$url/" | grep 6A34 | awk '{print $3}' | awk -F "-" '{print $2}' | awk -F "." '{print 20$3"-"$2"-"$1}')
t2=$(date +%s -d$t1)
now=$(date +%s)
echo $(((t2-now)/86400))

Is there a way to get the decimals?

Edit, it may be possible with use of awk and printf "%2f"
But not sure how.

Edit2: Solved
Code:
echo $t2 $now | awk '{printf "%.2f\n",($1-$2)/86400}'

Thanks to this forum, users and google Smilie

Last edited by Jotne; 01-28-2011 at 09:53 AM..
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DATETIMEIMMUTABLE.FORMAT(3)						 1					       DATETIMEIMMUTABLE.FORMAT(3)

DateTime::format - Returns date formatted according to given format

       Object oriented style

SYNOPSIS
public string DateTime::format (string $format) DESCRIPTION
string DateTimeImmutable::format (string $format) string DateTimeInterface::format (string $format) Procedural style string date_format (DateTimeInterface $object, string $format) Returns date formatted according to given format. PARAMETERS
o $object -Procedural style only: A DateTime object returned by date_create(3) o $format - Format accepted by date(3). RETURN VALUES
Returns the formatted date string on success or FALSE on failure. EXAMPLES
Example #1 DateTimeImmutable.format(3) example Object oriented style <?php $date = new DateTime('2000-01-01'); echo $date->format('Y-m-d H:i:s'); ?> Procedural style <?php $date = date_create('2000-01-01'); echo date_format($date, 'Y-m-d H:i:s'); ?> The above example will output: 2000-01-01 00:00:00 NOTES
This method does not use locales. All output is in English. SEE ALSO
date(3). PHP Documentation Group DATETIMEIMMUTABLE.FORMAT(3)