CODEWITHSUNDEEP
bashdatevariables
Josh
Josh

Reputation: 31

Using date command with multiple variables

I have three variables $year $month and $day. Each of them gather the input from a user like: myday 20160303 or myday 2016 03 03 in different ways, I used multiple "if" statements to save in this example the first yyyy under year variable, mm under month and day accordingly. I need to use them to find out the day of the week. What I have at this moment is

nn=`date --date="$1" +"%A"`

echo "$nn"

It works with input like 20160404 but if the user type 3 arguments how can I use all three variables correctly in a date command to show the day of the week?

Upvotes: 0

Views: 1431

Answers (2)

iamauser
iamauser

Reputation: 11489

This should work in linux date version (GNU coreutils), not sure about Mac OS X. date.bash looks like this:

#!/bin/bash
date -d "$1$2$3" +%Y-%m-%d

assign exe permission :

chmod u+x date.bash

execute:

]# ./date.bash 2016 06 24
2016-06-24

]# ./date.bash 20160624
2016-06-24

]#./date.bash 2016 0624
2016-06-24

For Mac OS X, you can use the following syntax in date.bash

#!/bin/bash
date -j -f "%Y%m%d" "$1$2$3" +%Y-%m-%d

In all the above examples, I used +%Y-%m-%d for the format. You can replace that with +%A as you are doing to get the name of the day.

Note: For Mac OS, ./date.bash 2016-06-24 will not work, while in Linux it would work.

Upvotes: 3

John Zwinck
John Zwinck

Reputation: 249582

If you want to combine three arguments into one, it's simply:

"$1$2$3"

This will turn 2016 03 03 into 20160303

Upvotes: 2

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