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[xsl] Find second character in a string (2)


Subject: [xsl] Find second character in a string (2)
From: "Paulo Henrique S. Bermejo" <bermejo@xxxxxxxxxxx>
Date: Fri, 31 Aug 2001 11:27:31 -0300

Personal,

To be more explicit, I write other once.
I'm using this below script:

<xsl:variable name="var-nome-preparado" select="'Clark Jenier Foz'"/>
<xsl:value-of
select="substring-before(substring-after(substring-after($var-nome-preparado
, ' '), ' '), ' ')"/>

And I only need to get the string "Clark Jenier".
If someone have a solution, advice me.

Thanks,

Paulo

----- Original Message -----
From: Paulo Henrique S. Bermejo
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Sent: Friday, August 31, 2001 11:00 AM
Subject: Find second character in a string


Hi Everybody,

Someone know if that function "contains" permitted to make to ask or to
especified if character repeat in a string?
Example:

I can know if repeat caracter ' ' (espace) in a string (Clark Jenier Foz)
this example exist 2 spaces in a string.
I need to make this because, I am making a script that get that last name
(in Brazil - "Foz") and converted to upper.

  <xsl:choose>
   <xsl:when test="contains(substring-after($var-nome-preparado, ' '), '
')">
    <xsl:call-template name="upper-texto">
     <xsl:with-param name="texto"
select="substring-after(substring-after($var-nome-preparado, ' '), ' ')"/>
    </xsl:call-template>
    <xsl:text>, </xsl:text>
    <xsl:value-of
select="substring-before(substring-after(substring-after($var-nome-preparado
, ' '), ' '), ' ')"/>
   </xsl:when>
  </xsl:choose>

When a run this example it return me only string "FOZ" because
"    <xsl:value-of
select="substring-before(substring-after(substring-after($var-nome-preparado
, ' '), ' '), ' ')"/> "

I see a ease resolution this, using scripts but I don't can to use.


Thanks for help!

Paulo Henrique Bermejo
Florianópolis/SC - Brazil.


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