Date Format in XSLT

Hello,

Checked it out but the code, but didnt work, pretty sure Im missing something. Please see below (see a Ja// instead).

<xsl:template match="/">
     <xsl:call-template name="showDateDropdown" />
  </xsl:template>

<xsl:template name="showDateDropdown">
    <xsl:call-template name="format-date">
      <xsl:with-param name="date" select="date" />
      <xsl:with-param name="format" select="1" />
    </xsl:call-template>
    <form name='drop2'>
      <select style="width:180px; background:#ccc; color:#000; font-weight:bold; font-size:10px;">
        <xsl:attribute name="onChange">
          <xsl:text>this.form.submit();</xsl:text>
        </xsl:attribute>
        <option value='0'>Select Date</option>
        <xsl:for-each select="//item[generate-id() = generate-id(key('item-by-date', date) [1])]">
          <option value='{date}'>
            <xsl:value-of select="date"/>
          </option>
        </xsl:for-each>
      </select>
    </form>
  </xsl:template>


 <!-- 1/8/04  01.08.04   Jan 8 2004 -->
  <xsl:template name="format-date">
    <xsl:param name="date" />
    <xsl:param name="format" select="0" />

    <xsl:variable name="day" select="substring-before(substring-after($date, ' '), ' ')" />
    <xsl:variable name="monthName" select="substring-before(substring-after(substring-after($date, ' '), ' '), ' ')" />
    <xsl:variable name="year" select="substring-before(substring-after(substring-after(substring-after($date, ' '), ' '), ' '), ' ')" />

    <xsl:variable name="month" select="substring(substring-after('Jan01Feb02Mar03Apr04May05Jun06Jul07Aug08Sep09Oct10Nov11Dec12', $monthName), 1, 2)" />
    <xsl:variable name="day2" select="concat(translate(substring($day,1,1), '0', ''), substring($day,2,1))" />
    <xsl:variable name="month2" select="concat(translate(substring($month,1,1), '0', ''), substring($month,2,1))" />

    <xsl:choose>
      <xsl:when test="$format = 1">
        <xsl:value-of select="concat($month2, '/', $day2, '/', substring($year, 3))" />
      </xsl:when>
      <xsl:when test="$format = 2">
        <xsl:value-of select="concat($month, '.', $day, '.', substring($year, 3))" />
      </xsl:when>
      <xsl:when test="$format = 3">
        <xsl:value-of select="concat($monthName, ' ', $day2, ' ', $year)" />
      </xsl:when>
      <xsl:when test="$format = 4">
        <xsl:value-of select="concat($day2, '/', $month2, '/', substring($year, 3))" />
      </xsl:when>

      <xsl:otherwise>
        <xsl:value-of select="$date" />
      </xsl:otherwise>
    </xsl:choose>
  </xsl:template>

Sorry, let me rephrase that in Englist :-)

*** Checked itout but the code didnt work, pretty sure Im missing something. Please see below (see a Ja// instead) ***

well, the code does work (though it doesn't provide the format you need),
 but you pass it the wrong context.
In this limited example you need to pass the files context as well
        <xsl:call-template name="format-date">
            <xsl:with-param name="date" select="files/date" />

note that with
<xsl:template match="/">
     <xsl:call-template name="showDateDropdown" />
  </xsl:template>
you pass the current context... and date is not a direct child of the root

now about the format-date template itself
first I am not really fond of the long series of substring-after manipulations,
specially not when there is reuse (if you have to do the same calculation twice, store the result in a variable)
but this is a minor detail

the format1 is not the format you need,
if you want to use the date-string in a sort, you don' have to strip off the "0"s, you want them for good sorting
your question was "16-09-2007"
the format 1 gives you 9/16/07 which is useless for sorting
but if you want to use the date for sorting, you will need 2007-09-16
assuming that that is what you want, I rewrote the format-date template

It seems like you are going to do a lot of on-and-off conversions, so I stripped off the options
This format date template only does 2007-09-16 as an output

    <xsl:template name="format-date">
        <xsl:param name="date"></xsl:param>
        <xsl:variable name="month-to-mm">Jan01JAN01Feb02FEB02Mar03MAR03Apr04APR04May05MAY05Jun06JUN06Jul07JUL07Aug08AUG08Sep09SEP09Oct10OCT10Nov11NOV11Dec12DEC12</xsl:variable>
       
        <xsl:variable name="dayless" select="substring-after($date, ' ')"></xsl:variable>
        <xsl:variable name="day" select="substring-before($dayless, ' ')" />
        <xsl:variable name="ddless" select="substring-after($dayless, ' ')"></xsl:variable>
        <xsl:variable name="monthName" select="substring-before($ddless, ' ')" />
        <xsl:variable name="year" select="substring-before(substring-after($ddless, ' '), ' ')" />
       
        <xsl:variable name="month" select="substring(substring-after($month-to-mm, $monthName), 1, 2)" />
        <xsl:variable name="day2">
                <xsl:if test="string-length($day) = 1">
                    <xsl:text>0</xsl:text>
                </xsl:if>
            <xsl:value-of select="$day"/>
        </xsl:variable>
         <xsl:value-of select="concat($year, '-', $month, '-', $day2)" />
    </xsl:template>

this one only takes one parameter

finally you want the dates in the dropdown list I assume, so not before it

Looking at your dropdown
I see from which question you took the dropdown code (are you a colleague of merdelafuente?)
but it hardly makes sense in your example
you are by the way using a key you did not declare

If you really need unique dates in the dropdown, you have to make a key
but then I strongly suggest that you transform all the dates first, if you can.
Can you change your source XML, so that the dates are in a sortable format?
(by doing an identity transform together with the format-date template)
If you need to keep the original date we could always add an attribute, like this
<files>
  <date yyyymmdd="2007-09-16">Mon, 16 Sep 2007 17:54:26 +0000</date>
  <enclosure file="custom_001.wmv />
</files>
Adding the preformatted date in the source XML, one way or another,
will make live a lot easier in the sorting and dropdown stage

let me knwo any further requirements

Geert

Hello Geert

Ahh that makes sense. Yes I saw that dropdown example here from EE  (in which you were one of the responders) It happens that i am working off of an XML from a 3rd party that I cannot change. I will give your example a try and will let you know.

Thx!

please note that you can have two steps in working with the XML... one step that transforms the source dates to a sortable form
and another step building the dropdown and doing the sorting

cheers

Geert

Hello Gertone

Your example worked great! I was able to build the dropdown with the unique keys -- one last question before I close the question if you dont mind?

Where do I actually call the sort statement? The display dropdown template or the main template? Here's the code that im working off of:

...
       <xsl:for-each select="//item[generate-id() = generate-id(key('item-by-date', date) [1])]">
          <xsl:sort select="date" order="descending"/>
          <option>            
            <xsl:attribute name="value">
              <xsl:call-template name="format-date">
                <xsl:with-param name="date" select="date" />
                </xsl:call-template>          
             </xsl:attribute>
            <xsl:value-of select="title"/>
          </option>
        </xsl:for-each>

I really, really appreciate your help :o)

Wow. I'll see if I can sort with something else.

Thanks again Geert Im sure you'll see me around!