Hi R.Singh your code match my requirement but at when use awk Im not getting expected results, can you please help me here
When I enter awk comd its not existing from cmd prompt Where I need to exit forcefully
my OS details
Last edited by Don Cragun; 10-22-2015 at 03:13 AM..
Reason: Put sample output in bold CODE tags.
hi,
i have 2 dates in the form: '20080315120030' and '20080310140030'. i.e. YYYYMMDDHHMMSS.
i need a way of getting the difference between them using shell script.
any thoughts? (14 Replies)
Hi
I want to write a BASH script.
I have files updated every hour from two platforms.The file names are :
- Falcon_smsCounters_1200020708.log
- Canari_smsCounters_1200020708.log
1200 refers to hour twelve and 020708 to 02 july 2008.
Inside every file i have a list of parameters... (2 Replies)
I am using two shell scripts a.ksh and b.ksh
a.ksh
1. Sets the value
+++++++++++++++++
export USER1=abcd1
export PASSWORD=xyz
+++++++++++++++++
b.ksh
2. Second scripts calls sctipt a.ksh and uses the values set in a.ksh and pass to an executable demo... (2 Replies)
Hi,
I have a file containing timestamps (at micro-seconds granularity). It looks like the following:
06:49:42.383818
06:49:42.390190
06:49:42.392308
06:49:42.392712
06:49:42.393437
06:49:42.393960
06:49:42.402115
Now I need a sed/awk script to take the difference of two successive... (2 Replies)
Dear all,
I fancy that I'm pretty competent in ksh, but I have someone on HP-UX wanting me to script up a simple interface to handle user alterations rather than giving them high privileges to run up SAM. This is all fairly straightforward, but I'm stuck on an epoch date issue.
When we have... (6 Replies)
1. The problem statement, all variables and given/known data:
The problem i have is that i probably make a few mistake here in the code but don't know what it is and i try to get the date difference but don't know where to add the days_in_month function
2. Relevant commands, code,... (1 Reply)
I tried the below code to find difference between two dates. It works fine if the day of the month is 2-digit number. But it fails when we have a single-digit day of month(ex:1-9). my code is as below. please help me soon.
#!/usr/bin/perl -w
use strict;
use Time::Local;
... (2 Replies)
Hi Everyone,
We are having an issue with date and date -u in our AIX Systems.
We have checked environment variable TZ and /etc/environment and however, we could not rectify the difference.
>date
Thu Mar 19 22:31:40 IST 2015
>date -u
Thu Mar 19 17:01:44 GMT 2015
Any clue... (5 Replies)
Hi All,
I have a CSV file which is as below. Basically I need to take the year column in it and find if the year is >= 20152 . If that is then I should subtract all values by 6. In the below example in description I am having number mentioned as YYWW so I need to subtract those by -5. Whereever... (8 Replies)
Discussion started by: arunkumar_mca
8 Replies
LEARN ABOUT PHP
dateinterval.format
DATEINTERVAL.FORMAT(3) 1 DATEINTERVAL.FORMAT(3)DateInterval::format - Formats the interval
SYNOPSIS
public string DateInterval::format (string $format)
DESCRIPTION
Formats the interval.
PARAMETERS
o $format
-
The following characters are recognized in the $format parameter string. Each format character must be prefixed by a percent sign (
%).
+------------------+--------------------------------------+---+
|$format character | | |
| | | |
| | Description | |
| | | |
| | Example values | |
| | | |
+------------------+--------------------------------------+---+
| | | |
| % | | |
| | | |
| | Literal % | |
| | | |
| | | |
| | % | |
| | | |
| | | |
| Y | | |
| | | |
| | Years, numeric, at least 2 digits | |
| | with leading 0 | |
| | | |
| | | |
| | 01, 03 | |
| | | |
| | | |
| y | | |
| | | |
| | Years, numeric | |
| | | |
| | | |
| | 1, 3 | |
| | | |
| | | |
| M | | |
| | | |
| | Months, numeric, at least 2 digits | |
| | with leading 0 | |
| | | |
| | | |
| | 01, 03, 12 | |
| | | |
| | | |
| m | | |
| | | |
| | Months, numeric | |
| | | |
| | | |
| | 1, 3, 12 | |
| | | |
| | | |
| D | | |
| | | |
| | Days, numeric, at least 2 digits | |
| | with leading 0 | |
| | | |
| | | |
| | 01, 03, 31 | |
| | | |
| | | |
| d | | |
| | | |
| | Days, numeric | |
| | | |
| | | |
| | 1, 3, 31 | |
| | | |
| | | |
| a | | |
| | | |
| | Total number of days as a result of | |
| | a DateTime::diff or (unknown) other- | |
| | wise | |
| | | |
| | | |
| | 4, 18, 8123 | |
| | | |
| | | |
| H | | |
| | | |
| | Hours, numeric, at least 2 digits | |
| | with leading 0 | |
| | | |
| | | |
| | 01, 03, 23 | |
| | | |
| | | |
| h | | |
| | | |
| | Hours, numeric | |
| | | |
| | | |
| | 1, 3, 23 | |
| | | |
| | | |
| I | | |
| | | |
| | Minutes, numeric, at least 2 digits | |
| | with leading 0 | |
| | | |
| | | |
| | 01, 03, 59 | |
| | | |
| | | |
| i | | |
| | | |
| | Minutes, numeric | |
| | | |
| | | |
| | 1, 3, 59 | |
| | | |
| | | |
| S | | |
| | | |
| | Seconds, numeric, at least 2 digits | |
| | with leading 0 | |
| | | |
| | | |
| | 01, 03, 57 | |
| | | |
| | | |
| s | | |
| | | |
| | Seconds, numeric | |
| | | |
| | | |
| | 1, 3, 57 | |
| | | |
| | | |
| R | | |
| | | |
| | Sign " -" when negative, " +" when | |
| | positive | |
| | | |
| | | |
| | -, + | |
| | | |
| | | |
| r | | |
| | | |
| | Sign " -" when negative, empty when | |
| | positive | |
| | | |
| | | |
| | -, | |
| | | |
+------------------+--------------------------------------+---+
RETURN VALUES
Returns the formatted interval.
NOTES
Note
The DateInterval::format method does not recalculate carry over points in time strings nor in date segments. This is expected
because it is not possible to overflow values like "32 days" which could be interpreted as anything from "1 month and 4 days" to "1
month and 1 day".
EXAMPLES
Example #1
DateInterval example
<?php
$interval = new DateInterval('P2Y4DT6H8M');
echo $interval->format('%d days');
?>
The above example will output:
4 days
Example #2
DateInterval and carry over points
<?php
$interval = new DateInterval('P32D');
echo $interval->format('%d days');
?>
The above example will output:
32 days
Example #3
DateInterval and DateTime::diff with the %a and %d modifiers
<?php
$january = new DateTime('2010-01-01');
$february = new DateTime('2010-02-01');
$interval = $february->diff($january);
// %a will output the total number of days.
echo $interval->format('%a total days')."
";
// While %d will only output the number of days not already covered by the
// month.
echo $interval->format('%m month, %d days');
?>
The above example will output:
31 total days
1 month, 0 days
SEE ALSO
DateTime::diff.
PHP Documentation Group DATEINTERVAL.FORMAT(3)