Inserting a line break in a PDF generated from XSL FO using

Question

I am using XSL FO to generate a PDF file containing a table with information. One of these columns is a \"Description\" column. An example of a string that I am populating o

Answer 1

Generating strings containing escaped XML markup is seldom the right answer, but if that's what you have to work with, then for input like this:

<Description><![CDATA[This is an example Description.<br/>List item 1<br/>List item 2<br/>List item 3<br/>List item 4]]></Description>

if you're using XSLT 2.0, you can use xsl:analyze-string to get the empty fo:block that you originally wanted:

<xsl:template match="Description">
  <fo:block>
    <xsl:analyze-string select="." regex="&lt;br/>">
      <xsl:matching-substring>
        <fo:block />
      </xsl:matching-substring>
      <xsl:non-matching-substring>
        <xsl:value-of select="." />
      </xsl:non-matching-substring>
    </xsl:analyze-string>
  </fo:block>
</xsl:template>

but if you are using XSLT 2.0, you can more concisely use linefeed-treatment="preserve" as per @Daniel Haley and use replace() to insert the linefeeds:

<xsl:template match="Description">
  <fo:block linefeed-treatment="preserve">
    <xsl:value-of select="replace(., '&lt;br/>', '&#xA;')" />
  </fo:block>
</xsl:template>

If you are using XSLT 1.0, you can recurse your way through the string:

<xsl:template match="Description">
  <fo:block linefeed-treatment="preserve">
    <xsl:call-template name="replace-br" />
  </fo:block>
</xsl:template>

<xsl:template name="replace-br">
  <xsl:param name="text" select="." />

  <xsl:choose>
    <xsl:when test="not(contains($text, '&lt;br/>'))">
      <xsl:value-of select="$text" />
    </xsl:when>
    <xsl:otherwise>
      <xsl:value-of select="substring-before($text, '&lt;br/>')"/>
      <xsl:text>&#xA;</xsl:text> <!-- or <fo:block /> -->
      <xsl:call-template name="replace-br">
        <xsl:with-param name="text" select="substring-after($text, '&lt;br/>')"/>
      </xsl:call-template>
    </xsl:otherwise>
  </xsl:choose>
</xsl:template>

Answer 2

I had a text block that looks like this

<fo:table-cell display-align="right">
<fo:block font-size="40pt" text-align="right">
    <xsl:text> Text 1 </xsl:text>
    <fo:block> </fo:block>
    <xsl:text> Text2 </xsl:text>
    <fo:block> </fo:block>
    <xsl:text> Text 3</xsl:text>
</fo:block>

NB: note the empty

Answer 3

For XSLT 1.0 I'm using my XSLT Line-Break Template on GitHub.

For XSL-FO it supports

• Line breaks
• Line delimiters (vs Line breaks)
• Series of pointers in a row
• Ignore Pointer Repetitions (disable the Series of pointers in a row)
• Any string as a pointer to insert a break or a delimiter ("\n" is default)
• Line delimiters' height
• Default Line delimiter height from a current font size.
• Auto ignoring of the "\r" char when searching a break place.
• Added support for XSLT 2.0 for a seamless migration.
• something else...

For XSLT 2.0 and later consider to use approaches like

• XSLT 2.0 xsl:analyze-string (RegEx)
• XPath 2.0 tokenize + XSLT (RegEx)
• passing sequences as a template parameter (XSLT 2.0)
• and so on

Answer 4

Try using linefeed-treatment="preserve" and \n instead of <br> for a new line.

<fo:block linefeed-treatment="preserve" >
 <xsl:value-of select="Description" />
</fo:block>

Answer 5

You could also replace <br/> with 
 and add a linefeed-treatment="preserve" attribute to your <fo:block>.

Something like:

<fo:block linefeed-treatment="preserve">This is an example Description.&#xA;List item 1&#xA;List item 2&#xA;List item 3&#xA;List item 4</fo:block>

Edit

Some users may need to use \n instead of 
 depending on how they are creating the XML. See Retain the 
 during xml marshalling for more details.