Sponsored Content
Top Forums Shell Programming and Scripting Converting Month into Integer Post 302662325 by itkamaraj on Tuesday 26th of June 2012 11:57:04 AM
Old 06-26-2012
Code:
$ echo "Jan" | awk 'BEGIN{months="  JanFebMarAprMayJunJulAugSepOctNovDec"}{print index(months,$0)/3}' 
1
$ echo "Feb" | awk 'BEGIN{months="  JanFebMarAprMayJunJulAugSepOctNovDec"}{print index(months,$0)/3}' 
2
$ echo "Dec" | awk 'BEGIN{months="  JanFebMarAprMayJunJulAugSepOctNovDec"}{print index(months,$0)/3}' 
12

These 3 Users Gave Thanks to itkamaraj For This Post:
 

10 More Discussions You Might Find Interesting

1. Shell Programming and Scripting

command for converting string to integer

Hi ... I am trying to calculate the time needed for a command to execute.. but the resulting value is getting as string.. so i am not able to use "expr " command.. please help me to convert the value to integer so that i can proceed with my script.. Regards esham (1 Reply)
Discussion started by: esham
1 Replies

2. Shell Programming and Scripting

Converting a decimal into integer

Hi all, I'm trying to convert a decimal number into an integer number; I'm doing this: n=`echo |awk '{ print "'"$mem"'"*10}'` where the variable mem is equal to 3.7 I'd like to obtain 37, but the expression above gives me 30. Help please!!!! thx a lot (4 Replies)
Discussion started by: idro
4 Replies

3. UNIX for Dummies Questions & Answers

Converting a String variable into Integer

Hi, I am passing a variable to a unix function. However when I try to assign the value to another variable like typeset -i I_CACHE_VAL=$2 Is this because of String to Integer conversion? I get an error. Please help me with thsi. Thanks (2 Replies)
Discussion started by: neeto
2 Replies

4. Shell Programming and Scripting

Converting integer to String

Hi everyone, I would like to know how to convert an integer to a string. for instance if i=1 i would like to creat a variable called constant1. i want to do this in a for loop so for each value of i, i create a new variable such as constant2, constant3,... and so on. for i in 1 2 3 do ... (1 Reply)
Discussion started by: ROOZ
1 Replies

5. Shell Programming and Scripting

Converting decimal to integer

The shell mentioned below will show a warning if the page takes more than 6 seconds to load. The problem is that myduration variable is not an integer. How do I convert it to integer? myduration=$(curl http://192.168.50.1/mantisbt/view.php?id=1 -w %{time_total}) > /dev/null ; ] && echo... (3 Replies)
Discussion started by: shantanuo
3 Replies

6. Shell Programming and Scripting

Converting date +%j into integer

Dear community, i got a problem to get "date +%j" as the right value. Today is the 10th day of the year. #./script.sh 2 #!/bin/bash/ Var1=$(date +%j) Var2=$1 let result=$Var1+$Var2 echo $Var1 plus $Var2 equals $result The output of the script is: 010 plus 2 equals 10... (9 Replies)
Discussion started by: OskarHF
9 Replies

7. Shell Programming and Scripting

Converting string to integer

I have a function that is supposed to check for user processes and wait for 0 count before exiting the function. I am sure I have more than one issue in my code, but the stumbling block right now is that I am trying to convert the value of my variable from a string to integer. process_count... (10 Replies)
Discussion started by: MIA651
10 Replies

8. Shell Programming and Scripting

Converting String To Integer/Float (weird case)

Hi guys, I'm new here. I have a problem at work. One of our scripts was eventually having a bug and only detected recently. Here's the issue and background: Bash Script which calls AWK script Awk script returns a string as per below (example):var1='00000-123'So, when we convert it, the... (18 Replies)
Discussion started by: sekfarok
18 Replies

9. Shell Programming and Scripting

Converting time format as Integer to seconds

Hello, How can we convert date like format 20181004171050 in seconds ? I can able to convert till date but failing for HHMMSS. date -d "20181004" "+%s" output as 1538596800 . But when i add hhmmss it is failing date -d "20181004172000" "+%s" result Invalid date Kindly guide. Regards (16 Replies)
Discussion started by: sadique.manzar
16 Replies

10. UNIX for Beginners Questions & Answers

Replace integer string in a variable based on month?

Hi Folks - Linux Version = Linux 2.6.39-400.128.17.el5uek x86_64 I have a process that determines the start and end load periods for an Oracle data load process. The variables used are as follows follows: They are populated like such: However, the load requires the month to be the... (11 Replies)
Discussion started by: SIMMS7400
11 Replies
DATETIME.ADD(3) 							 1							   DATETIME.ADD(3)

DateTime::add - Adds an amount of days, months, years, hours, minutes and seconds to a DateTime object

       Object oriented style

SYNOPSIS
public DateTime DateTime::add (DateInterval $interval) DESCRIPTION
Procedural style DateTime date_add (DateTime $object, DateInterval $interval) Adds the specified DateInterval object to the specified DateTime object. PARAMETERS
o $object -Procedural style only: A DateTime object returned by date_create(3). The function modifies this object. o $interval - A DateInterval object RETURN VALUES
Returns the DateTime object for method chaining or FALSE on failure. EXAMPLES
Example #1 DateTime.add(3) example Object oriented style <?php $date = new DateTime('2000-01-01'); $date->add(new DateInterval('P10D')); echo $date->format('Y-m-d') . " "; ?> Procedural style <?php $date = date_create('2000-01-01'); date_add($date, date_interval_create_from_date_string('10 days')); echo date_format($date, 'Y-m-d'); ?> The above examples will output: 2000-01-11 Example #2 Further DateTime.add(3) examples <?php $date = new DateTime('2000-01-01'); $date->add(new DateInterval('PT10H30S')); echo $date->format('Y-m-d H:i:s') . " "; $date = new DateTime('2000-01-01'); $date->add(new DateInterval('P7Y5M4DT4H3M2S')); echo $date->format('Y-m-d H:i:s') . " "; ?> The above example will output: 2000-01-01 10:00:30 2007-06-05 04:03:02 Example #3 Beware when adding months <?php $date = new DateTime('2000-12-31'); $interval = new DateInterval('P1M'); $date->add($interval); echo $date->format('Y-m-d') . " "; $date->add($interval); echo $date->format('Y-m-d') . " "; ?> The above example will output: 2001-01-31 2001-03-03 NOTES
DateTime.modify(3) is an alternative when using PHP 5.2. SEE ALSO
DateTime.sub(3), DateTime.diff(3), DateTime.modify(3). PHP Documentation Group DATETIME.ADD(3)
All times are GMT -4. The time now is 05:59 AM.
Unix & Linux Forums Content Copyright 1993-2022. All Rights Reserved.
Privacy Policy