Question

我有一个包含XML值的变量

<xsl:variable name="resp">
<TABLE>
 <news_id>39</news_id>
 <news_title>Strada &#8211; TSO</news_title>
 <news_content>Extending the support to IFA&#8217;s family in form of income security by paying last drawn trail paid to IFA before the demise.</news_content>
 <news_type>1</news_type>
 <doc_url>upload/docs/12345.pdf</doc_url>
 <doc_size>185859</doc_size>
 <news_date>2014-09-19 12:54:11.0</news_date>
 <is_enabled>1</is_enabled>
</TABLE>  
<xsl:variable>

如何解析此变量中的值。请帮忙。

Answer 1

我要拥有这个结构，你这样做：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    exclude-result-prefixes="xs"
    version="1.0">
    <xsl:output indent="yes"/>
    <xsl:template match="/">
        <xsl:variable name="resp">
            <TABLE>
                <news_id>39</news_id>
                <news_title>Strada &#8211; TSO</news_title>
                <news_content>Extending the support to IFA&#8217;s family in form of income security by paying last drawn trail paid to IFA before the demise.</news_content>
                <news_type>1</news_type>
                <doc_url>upload/docs/12345.pdf</doc_url>
                <doc_size>185859</doc_size>
                <news_date>2014-09-19 12:54:11.0</news_date>
                <is_enabled>1</is_enabled>
            </TABLE>  
        </xsl:variable>

        <!-- using copy-of to get the structure of the XML -->
        <xsl:copy-of select="$resp" />
    </xsl:template>
</xsl:stylesheet>

如果您想要变量中的文本，请执行以下操作：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    exclude-result-prefixes="xs"
    version="1.0">
    <xsl:output indent="yes"/>
    <xsl:template match="/">
        <xsl:variable name="resp">
            <TABLE>
                <news_id>39</news_id>
                <news_title>Strada &#8211; TSO</news_title>
                <news_content>Extending the support to IFA&#8217;s family in form of income security by paying last drawn trail paid to IFA before the demise.</news_content>
                <news_type>1</news_type>
                <doc_url>upload/docs/12345.pdf</doc_url>
                <doc_size>185859</doc_size>
                <news_date>2014-09-19 12:54:11.0</news_date>
                <is_enabled>1</is_enabled>
            </TABLE>  
        </xsl:variable>
        <variable>
        <!-- using value-of to get the values of the XML inserting it into variable element -->
            <xsl:value-of select="$resp"/>
        </variable>


    </xsl:template>
</xsl:stylesheet>

如果您想从变量中的元素中获取不同的值，请执行以下操作：

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    exclude-result-prefixes="xs"
    version="1.0">
    <xsl:output indent="yes"/>
    <xsl:template match="/">
        <xsl:variable name="resp">
            <TABLE>
                <news_id>39</news_id>
                <news_title>Strada &#8211; TSO</news_title>
                <news_content>Extending the support to IFA&#8217;s family in form of income security by paying last drawn trail paid to IFA before the demise.</news_content>
                <news_type>1</news_type>
                <doc_url>upload/docs/12345.pdf</doc_url>
                <doc_size>185859</doc_size>
                <news_date>2014-09-19 12:54:11.0</news_date>
                <is_enabled>1</is_enabled>
            </TABLE>  
        </xsl:variable>

        <!-- If you want to get the value of news_id inside the variable you do this -->
        <newsId>
            <xsl:value-of select="$resp/TABLE/news_id"/>
        </newsId>

        <!-- Below is a shorter xpath-syntax -->
        <newsTitle>
            <xsl:value-of select="$resp//news_title"/>
        </newsTitle>

    </xsl:template>
</xsl:stylesheet>

希望这有帮助。

转换＆amp;到＆amp;在xslt

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