使用 XSLT 合并两个 XML 文件
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Merging two XML files using XSLT
提问by user1457881
I have 2 xml files which I need to merge together using a style sheet
我有 2 个 xml 文件，我需要使用样式表将它们合并在一起
<AssessmentInput>
<ApplicationData>...</AppicationData>
<MetricList>...</MetricList>
</AssessmentInput>
One is ApplicationData and the other one is MetricList. here is what I have done but it is nothing close to what it should be
一个是ApplicationData，另一个是MetricList。这是我所做的，但与它应该做的完全不同
<?xml version="1.0" encoding="ascii"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:exslt="http://exslt.org/common" exclude-result-prefixes="xsl exslt">
<xsl:output omit-xml-declaration="yes" method="xml" encoding="UTF-8"/>
<xsl:param name="ApplicationData" select="/"/>
<xsl:param name="MetricList"/>
<xsl:template match="/">
<xsl:apply-templates select="$ApplicationData/ApplicationData"/>
</xsl:template>
<xsl:template match="ApplicationData">
<xsl:copy>
<xsl:apply-templates select="@*"/>
<xsl:apply-templates select="*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>
Please help me. I don't have any experience with XSLT.
请帮我。我对 XSLT 没有任何经验。
回答by hielsnoppe
Given the following input files:
给定以下输入文件：
ApplicationData.xml
应用数据.xml
<?xml version="1.0" ?>
<ApplicationData>
Whatever data you have in here.
</ApplicationData>
MetricList.xml
指标列表.xml
<?xml version="1.0" ?>
<MetricList>
Whatever list you have in here.
</MetricList>
AssessmentInput.xml
评估输入文件
<?xml version="1.0" ?>
<AssessmentInput />
the following transformation merge.xslapplied to AssessmentInput.xml
以下转换merge.xsl应用于AssessmentInput.xml
<?xml version="1.0" ?>
<xsl:transform
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
version="1.0">
<xsl:template match="/AssessmentInput">
<xsl:copy>
<xsl:copy-of select="document('ApplicationData.xml')" />
<xsl:copy-of select="document('MetricList.xml')" />
</xsl:copy>
</xsl:template>
</xsl:transform>
produces the correct output of
产生正确的输出
<?xml version="1.0" encoding="UTF-8"?>
<AssessmentInput>
<ApplicationData>
Whatever data you have in here.
</ApplicationData>
<MetricList>
Whatever list you have in here.
</MetricList>
</AssessmentInput>
回答by kaushik0033
You should be replace your line as below:-
您应该按如下方式替换您的线路：-
<xsl:param name="ApplicationData" select="/"/>
<xsl:param name="MetricList"/>
with this below line one:
使用下面的第一行：
<xsl:variable name="Application_Data" select="document('Application_Data.xml')/ApplicationData"/>
<xsl:variable name="'Metric_List" select="document('Application_Data.xml')/MetricList"/>
i think this may help you..
我想这可能会帮助你..
