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. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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