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up vote 0 down vote favorite

My shell script:

#!/bin/bash
echo "$1";
startd=$(date -d "$1" +"%Y%m%d"); 
echo "$startd";

My command:

sudo ./test.sh "20151010"

The output:

20151010 
20150213

it printed todays date instead of printing the input date any idea?

share|improve this question
    
Be warned: date -d with sudo might change system date in OSX –  Ketan Feb 13 at 23:12
    
I have no such problem with date under Linux. You may need to install the GNU Coreutils and use date from them. BTW, you do not need sudo. –  vinc17 Feb 13 at 23:15
1  
The OS X version of date doesn't parse arbitrary date formats like the GNU version does. –  Barmar Feb 13 at 23:23

2 Answers 2

up vote 2 down vote accepted

The OS X version of date uses the -f option to parse a formatted date/time:

date -j -f '%Y%m%d' "$1" +'%Y%m%d'

The -j option causes it to just print the time, not try to set the system clock. So the full script would be:

#!/bin/bash
echo "$1";
startd=$(date -j -f '%Y%m%d' "$1" +'%Y%m%d'); 
echo "$startd";

Here's a transcript:

$ ./testdate.sh 20151010
20151010
20151010
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it turns out developing these scripts on osx won't do for me than because production env is linux, And i want the linux version of scripts –  jaminator Feb 17 at 19:22
up vote 1 down vote

If you are trying to format date on OS X, you can try this:

date -j -f "%Y%m%d" "20151010"

I get the following output:

Sat Oct 10 17:27:28 CDT 2015
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