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Today's date minus input number

Matthew Singleton

So I have a macro that will subtract a number of days from today's date, is there a way of having an input that determines the numbers of days to be subtracted i.e Macro(5) would subtract 5 days from today, macro(12) would be todays date minus 12 days etc.

here is the hardcoded date marco:

#set ( $date = $action.dateFormatter.calendar )
$date.setTime($content.currentDate)
$date.add(6, -3)
<time datetime="$action.dateFormatter.formatGivenString('yyyy-MM-dd', $date.time)"/>

Accepted answers

Christo Mastoroudes [Adaptavist]

Hi,

I assume you are using a user macro.

Yes, you will have to specify a macro parameter of type "int" and make it "required" so that the user must specify it.

Then you can reference it in your template code and make the appropriate calculation.

Screen Shot 2018-01-31 at 11.56.27.png Documentation: https://confluence.atlassian.com/doc/writing-user-macros-4485.html

If this answers your question please mark it as answered.

All comments

Christo Mastoroudes [Adaptavist]

Hi,

I assume you are using a user macro.

Yes, you will have to specify a macro parameter of type "int" and make it "required" so that the user must specify it.

Then you can reference it in your template code and make the appropriate calculation.

Screen Shot 2018-01-31 at 11.56.27.png Documentation: https://confluence.atlassian.com/doc/writing-user-macros-4485.html

If this answers your question please mark it as answered.

Matthew Singleton

Thanks but it seems to not like a variable in $date.add as I get:

$date.add(6, $paramName) 31 Jan 2018

when using this:

## @param Name:title=Name|type=int|default=12|required=true

#set ( $date = $action.dateFormatter.calendar )
$date.setTime($content.currentDate)
$date.add(6, $paramName)
<time datetime="$action.dateFormatter.formatGivenString('yyyy-MM-dd', $date.time)"/>

Christo Mastoroudes [Adaptavist]

Try just to print out the new variable directly on the page before you use it in a function to ensure that it has been declared correctly.

Matthew Singleton

Yep that works fine!

If I use

...

$date.add(6, $paramName)
<time datetime="$action.dateFormatter.formatGivenString('yyyy-MM-dd', $date.time)"/>
$paramName

I get:

$date.add(6, $paramName) 31 Jan 2018 12

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