質問

I would like a tip, please :

To convert a String to Date in velocity:

It is $convert.parseDate($currentMessage.date.begin)

I try too :

#set($str = $!currentMessage.date.begin)
$str
#set($dateTransforme = $date.toDate('yyyy-MM-dd', $date.date))
$dateTransforme 

$dateTransforme.parseDate($str)  <br />

N.B. $currentMessage.date.begin is a string.

I continue to have at the running :

$dateTransforme.parseDate($str)

Why? My string is in the format '2014-02-26'

Thanks,

役に立ちましたか?

解決

Here is the working code with datetool

#set ($toDateRegDate= $dateTool.toDate('yyyy-MM-dd hh:mm:ss Z', $Video.Date.getData()))
#set ($videoDate = $dateTool.format('MMMM dd,yyyy', $toDateRegDate,$locale))

$videoDate 

他のヒント

Finally, It's OK:

  ## tranformation of the data to dates : date.begin
  #set($dateBegin = $date.toDate('yyyy-MM-dd',$!currentMessage.date.begin))

  ## tranformation of the data to dates :  date.end
  #set($dateEnd = $date.toDate('yyyy-MM-dd',$!currentMessage.date.end))

Ale

Convert a string to a date object

#set($dateObj = $date.toDate("dd/MM/yyyy", "08/09/2015"))

Format the date objet to "yyyy-MM-dd"

#set( $dateFormated = $date.format("yyyy-MM-dd", $dateObj))
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