Come posso tagliare lo spazio in XSLT senza sostituire gli spazi bianchi di replicazione da quelli singoli?

Question

La funzione normalize-space sostituisce le sequenze di spazi bianchi con un singolo spazio e taglia la stringa fornita. Come posso tagliare solo la stringa senza la sostituzione di spazi bianchi? Esistono soluzioni proprietarie come orcl:left-trim, ma sto cercando uno non proprietario.

Esempio:

<xsl:value-of select="trim(/Car/Description)"/> 

dovrebbe girare

<car> 
    <description> To get more information look at:  www.example.com </description> 
</car> 

in

"To get more information look at:  www.example.com" 

Answer 1

Una soluzione che utilizza solo xslt 1.0 modelli: codice

<xsl:variable name="whitespace" select="'&#09;&#10;&#13; '" /> 

<!-- Strips trailing whitespace characters from 'string' --> 
<xsl:template name="string-rtrim"> 
    <xsl:param name="string" /> 
    <xsl:param name="trim" select="$whitespace" /> 

    <xsl:variable name="length" select="string-length($string)" /> 

    <xsl:if test="$length &gt; 0"> 
     <xsl:choose> 
      <xsl:when test="contains($trim, substring($string, $length, 1))"> 
       <xsl:call-template name="string-rtrim"> 
        <xsl:with-param name="string" select="substring($string, 1, $length - 1)" /> 
        <xsl:with-param name="trim" select="$trim" /> 
       </xsl:call-template> 
      </xsl:when> 
      <xsl:otherwise> 
       <xsl:value-of select="$string" /> 
      </xsl:otherwise> 
     </xsl:choose> 
    </xsl:if> 
</xsl:template> 

<!-- Strips leading whitespace characters from 'string' --> 
<xsl:template name="string-ltrim"> 
    <xsl:param name="string" /> 
    <xsl:param name="trim" select="$whitespace" /> 

    <xsl:if test="string-length($string) &gt; 0"> 
     <xsl:choose> 
      <xsl:when test="contains($trim, substring($string, 1, 1))"> 
       <xsl:call-template name="string-ltrim"> 
        <xsl:with-param name="string" select="substring($string, 2)" /> 
        <xsl:with-param name="trim" select="$trim" /> 
       </xsl:call-template> 
      </xsl:when> 
      <xsl:otherwise> 
       <xsl:value-of select="$string" /> 
      </xsl:otherwise> 
     </xsl:choose> 
    </xsl:if> 
</xsl:template> 

<!-- Strips leading and trailing whitespace characters from 'string' --> 
<xsl:template name="string-trim"> 
    <xsl:param name="string" /> 
    <xsl:param name="trim" select="$whitespace" /> 
    <xsl:call-template name="string-rtrim"> 
     <xsl:with-param name="string"> 
      <xsl:call-template name="string-ltrim"> 
       <xsl:with-param name="string" select="$string" /> 
       <xsl:with-param name="trim" select="$trim" /> 
      </xsl:call-template> 
     </xsl:with-param> 
     <xsl:with-param name="trim" select="$trim" /> 
    </xsl:call-template> 
</xsl:template> 

prova:

<ltrim> 
    <xsl:call-template name="string-ltrim"> 
     <xsl:with-param name="string" select="' &#10; test '" /> 
    </xsl:call-template> 
</ltrim> 
<rtrim> 
    <xsl:call-template name="string-rtrim"> 
     <xsl:with-param name="string" select="' &#10; test &#10; '" /> 
    </xsl:call-template> 
</rtrim> 
<trim> 
    <xsl:call-template name="string-trim"> 
     <xsl:with-param name="string" select="' &#10; test &#10; '" /> 
    </xsl:call-template> 
</trim> 

uscita:

<test> 
    <ltrim>test </ltrim> 
    <rtrim> 
    test</rtrim> 
    <trim>test</trim> 
</test> 

Answer 2

Utilizzando FXSL (libreria open source per XSLT programmazione funzionale, scritto interamente in XSL T) si scrive semplicemente:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
    <xsl:import href="trim.xsl"/> 

    <xsl:output method="text"/> 
    <xsl:template match="/*/description"> 
    '<xsl:call-template name="trim"> 
     <xsl:with-param name="pStr" select="."/> 
    </xsl:call-template>' 
    </xsl:template> 
</xsl:stylesheet> 

Quando questa trasformazione viene applicata sul documento XML fornito:

<car> 
    <description> To get more information look at:  www.example.com </description> 
</car> 

The Wanted, risultato corretto è prodotto:

'To get more information look at:  www.example.com' 

---

Come funziona il modello trim?

Taglia gli spazi bianchi iniziali, quindi inverte la stringa risultante e taglia gli spazi bianchi iniziali, quindi inverte infine la stringa risultante.

---

II. XPath soluzione 2,0:

Uso:

replace(replace(/*/description, '^\s*(.+?)\s*$', '$1'), '^ .*$', '') 

Ecco un XSLT - 2.0 - verifica base:

<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> 
    <xsl:output omit-xml-declaration="yes" indent="yes"/> 

<xsl:template match="/"> 
    "<xsl:sequence 
     select="replace(replace(/*/description, '^\s*(.+?)\s*$', '$1'), '^ .*$', '')"/>" 
</xsl:template> 
</xsl:stylesheet> 

Quando questa trasformazione viene applicata sul documento XML fornito (sopra), l'espressione XPath viene valutata e il risultato di questa valutazione viene copiato nell'output:

"To get more information look at:  www.example.com" 

Answer 3

Se non si dispone di eventuali spazi in mezzo, si può semplicemente utilizzare:

translate(/Car/Description,' ','') 

Answer 4

normalize-space (actualSting) - Questo lo farà.

Answer 5

A breve soluzione XSLT1:

<xsl:template name="trim"> 
      <xsl:param name="str"/> 

      <xsl:choose> 
       <xsl:when test="string-length($str) &gt; 0 and substring($str, 1, 1) = ' '"> 
        <xsl:call-template name="trim"><xsl:with-param name="str"><xsl:value-of select="substring($str, 2)"/></xsl:with-param></xsl:call-template></xsl:when> 
       <xsl:when test="string-length($str) &gt; 0 and substring($str, string-length($str)) = ' '"> 
        <xsl:call-template name="trim"><xsl:with-param name="str"><xsl:value-of select="substring($str, 1, string-length($str)-1)"/></xsl:with-param></xsl:call-template></xsl:when> 
       <xsl:otherwise><xsl:value-of select="$str"/></xsl:otherwise> 
      </xsl:choose> 
     </xsl:template>