i have added the below xsl:output tag in xslt
<xsl:output method="html" indent="yes" encoding="utf-8" doctype-public="-//W3C//DTD HTML 4.0 Transitional//EN" ></xsl:output>
as a result i get the below doctype tag in the html output-
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
how can i开发者_JAVA技巧 mention the url in the doctype tag using xsl:output which would output a doctype tag that looks like below
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "_http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
Best Regards,
KeshavA quick trip to W3Schools reveals that the XSL:Output element has a doctype-system attribute available, as well as a doctype-public. I believe this is what you need.
If you change your xsl:output to the following
<xsl:output method="html" indent="yes" encoding="utf-8" doctype-public="-//W3C//DTD HTML 4.0 Transitional//EN" doctype-system="_http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd" ></xsl:output>
You should get the DTD output you require, as follows:
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN" "_http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
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