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Pipe with extra argument

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# 1  
Old 06-21-2011
Pipe with extra argument
Hi,

I want to pipe an output to another command with an extra argument. Can anyone tell me how is it possible?

Eg:
Code:
$date +%d

will give the current date. I want to add 1 to it and give it as argument to my calc command.
Code:
$calc "todays date" + 1

Basically i want to diplay tomorrows date. Any method for that?

I tried doing this
Code:
$ calc | date +%d

The calc command is nothing but the following:
Code:
#!/bin/bash
echo "scale=4; $1" | bc ;exit

Last edited by Franklin52; 06-21-2011 at 03:03 AM.. Reason: Please use code tags for code and data samples
# 2  
Old 06-21-2011
do you have gnu date ?

execute the below commands and check

Code:
 
date --date='tomorrow' +%d
date --date='1 day' +%d

# 3  
Old 06-21-2011
I am using Sun OS.
Those commands dont work. It says illegal options
# 4  
Old 06-21-2011
Then you are out of luck. Can you use Perl or Python (and what versions)?
# 5  
Old 06-21-2011
Hi
Check this perderabo's date script

Guru.
# 6  
Old 06-21-2011
Code:
 
bash-3.00$ perl -e 'print scalar(localtime(time + 86400)), "\n"'
Wed Jun 22 07:08:43 2011
bash-3.00$ date
Tue Jun 21 07:08:45 BST 2011
bash-3.00$

# 7  
Old 06-21-2011
@itkamaraj
That was awesome. It works fine.
Can you tell me how to extract only the day or month from it?
Like
Code:
$date +%d

will give you only the current day.
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