On Nov 7, 2007, at 16:20, Andreas Siepert wrote:
Hi
Isn't there another way?
Sorry, I don't think so...
The only way to get a guaranteed linebreak is either:
* start a new fo:block
* specify linefeed-treatment='preserve' on the containing block, and
use linefeed characters
Like you suggested, I would have to write my fo stuff into a single
line, otherwise there will be lots of additional linefeeds.
Well, this is generally how serialized FO reaches FOP (no indents).
If you use XSLT to generate the FO, specify no indent="yes" on the
xsl:output element and do not use xsl:preserve-space in your
stylesheet, then any compliant XSLT processor should output an XML
document that has all content on one line (unless, of course the
input contains text() nodes with 'significant' linefeeds...)
If you use XSLT, it would thus not really matter, since this very
human-readable:
<xsl:template match="somenode">
<fo:block linefeed-treatment="preserve">
<fo:inline>
<xsl:value-of select="othernode" />
<xsl:text>
and a linefeed</xsl:text>
</fo:inline>
More text
</fo:block>
</xsl:template>
Causes the following, very processor-friendly FO output (if using no
special output options, and using default XSLT whitespace stripping):
<fo:block linefeed-treatment="preserve"><fo:inline>{content of
othernode, possibly including linefeeds that will be preserved}

and a linefeed</fo:inline>More text</fo:block>
Note: The above should be a single line. Any linebreaks appearing,
apart from the numeric character reference, will be a result of line-
wrapping elsewhere.
Cheers
Andreas
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