XSL 查询内部 xml [英] XSL query inner xml

问题描述

说这是我的 xml :

Say this is my xml :

<History>
    <University>TSU</University>
    <Payload>
        <Attrib Order="0">OVERSEA</Attrib>
        <Attrib Order="1">GRADE2</Attrib>
        <Attrib Order="2">&lt;Person&gt;&lt;ID&gt;TQR344&lt;/ID&gt;&lt;/Person&gt;</Attrib>
        <Attrib Order="3">3566644</Attrib>
    </Payload>
</History>

我想查询 Order=2 标签内的内部 XML 并读取此人的 ID.

And I want to query the inner XML inside Order=2 tag and read ID of the person.

到目前为止我已经创建了这个:

I have created this so far :

<xsl:stylesheet version="1.0"
     xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
     >

    <xsl:output method="xml" encoding="UTF-8" indent="yes" omit-xml-declaration="no" />

    <xsl:template match="/History">
            <xsl:apply-templates select="/History" />
    </xsl:template>

    <xsl:template name="Person" match="//History">
        <Event>
            <Uni><xsl:value-of select="University" /></Uni>
            <ID><xsl:value-of select="Payload/Attrib[@Order='2']/Person/ID"  disable-output-escaping="yes" /></ID>
        </Event>
    </xsl:template>
</xsl:stylesheet>

但是正如你所看到的，它不起作用.

But as you can see it is not working.

我还将内部 XML 分配给一个变量并尝试查询该变量，但它也不起作用.

Also I assigned the inner XML into a variable and tried to query that variable and It didn't work too.

是否可以通过 xsl 做到这一点?

Is it possible to do that via xsl ?

限制:我无法更改 xml 格式.但也许我能够从 xsl ver 1 迁移到新版本.

Limitations : I cannot change xml format. But maybe I was able to move from xsl ver 1 to new versions.

推荐答案

我想查询 Order=2 标签内的内部 XML

I want to query the inner XML inside Order=2 tag

有问题的标签不包含任何 XML；它的内容是一个字符串，需要使用字符串函数进行操作.试试:

The tag in question does not contain any XML; its content is a string and needs to be manipulated using string functions. Try:

XSLT 1.0

<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="utf-8" indent="yes"/>

<xsl:template match="/History">
    <Event>
        <Uni>
            <xsl:value-of select="University" />
        </Uni>
        <ID>
            <xsl:value-of select="substring-before(substring-after(Payload/Attrib[@Order='2'], '&lt;ID&gt;'),'&lt;/ID&gt;&lt;')"/>
        </ID>
    </Event>
</xsl:template>

</xsl:stylesheet>

注意:

1.这:

<xsl:template match="/History">
    <xsl:apply-templates select="/History" />
</xsl:template>

创建一个无限循环并会使您的处理器崩溃.

creates an infinite loop and will crash your processor.

2. 或者，您可以将字符串序列化回 XML 并将结果作为 XML 处理；在 XSLT 1.0 中，这只能通过输出禁用转义的字符串，将结果保存为新文档，然后使用另一个 XSLT 样式表处理新文档来完成.使用 XSLT 3.0(或支持序列化作为扩展的处理器)，这一切都可以在同一个转换期间完成.

2. Alternatively, you could serialize the string back into XML and process the result as XML; in XSLT 1.0, this can be done only by outputting the string with the escaping disabled, saving the result as a new document, then processing the new document with another XSLT stylesheet. Using XSLT 3.0 (or a processor that supports serializing as an extension) this can be all done during the same transformation.

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