Question

交叉发布http://perlmonks.org/index.pl?node_id=979710

我正在尝试使用Perl和Lib :: XSLT从某些XML创建一个文本文件，我的转换工作正常，除了Lib :: XSLT将不需要的？xml版本标记添加到文件的开头，我该如何停止它这样做？

这是我的XSLT：

<xslt:stylesheet version="1.0" xmlns:data="http://www.SDMX.org/resources/SDMXML/schemas/v2_0/generic" xmlns:xslt="http://www.w3.org/1999/XSL/Transform" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:message="http://www.SDMX.org/resources/SDMXML/schemas/v2_0/message" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text" omit-xml-declaration="yes"/>
<xslt:param name="sep">|</xslt:param>
<xslt:output method="text" />
<xslt:template match="message:MessageGroup">
<xslt:for-each select="data:DataSet">
<!-- get dimensions (but not time) and store in dimensions variable -->
<xslt:for-each select="data:Series">
<xslt:variable name="dimensions">
<xslt:for-each select="data:SeriesKey">
<xslt:for-each select="data:Value">
<xslt:value-of select="@value" />
<xslt:value-of select="$sep" />
</xslt:for-each>
</xslt:for-each>
</xslt:variable>
<!--get obs statuses and store in obs statuses variable-->
<xslt:variable name="obsStatuses">
<xslt:for-each select="data:Attributes">
<xslt:for-each select="data:Value">
<xslt:value-of select="@value" />
</xslt:for-each>
</xslt:for-each>
</xslt:variable>
<!--write out dimensions variable, time, observation, obsstatuses variable-->
<xslt:for-each select="data:Obs">
<xslt:value-of select="$dimensions" />
<xslt:value-of select="data:Time" />
<xslt:value-of select="$sep" />
<xslt:value-of select="data:ObsValue/@value" />
<xslt:value-of select="$sep" />
<xslt:value-of select="data:Attributes/data:Value/@value"/>
<xslt:text>
</xslt:text>
</xslt:for-each>
</xslt:for-each>
</xslt:for-each>
</xslt:template>
</xslt:stylesheet>

这是Perl：

use Lib::XSLT;
my $parser = XML::LibXML->new();
my $xslt = XML::LibXSLT->new();
my $source = XML::LibXML->load_xml(location => "$xmlFile");
my $style_doc = $parser->parse_file(Path::Class::File->new("$xsltFile"));
my $stylesheet = $xslt->parse_stylesheet($style_doc);
open OUTPUTFILE, ">>$outputFile" or die("Unable to open $outputFile, $!");
print OUTPUTFILE $stylesheet->transform($source);
close OUTPUTFILE;

Answer 1

存储$ stylesheet-＆gt; transform（）的结果并使用$ stylesheet-＆gt; output_file（）修复了这个问题，例如：

use Lib::XSLT;
my $parser = XML::LibXML->new();
my $xslt = XML::LibXSLT->new();
my $source = XML::LibXML->load_xml(location => "$xmlFile");
my $style_doc = $parser->parse_file(Path::Class::File->new("$xsltFile"));
my $stylesheet = $xslt->parse_stylesheet($style_doc);
my $results = $stylesheet->transform($source);
$stylesheet->output_file($results, $outputFile);

Answer 2

为什么<?xml>声明不受欢迎？它是有效的XML，并且对解析器没有影响。

使用Lib：XSLT创建文本文件时的额外xml标记

2 个答案: