用 XSLT 替换节点文本? [英] Replacing the Node text with XSLT?
问题描述
下面是我的XML文件，用于存储数据-
Below is my XML file, which is used to stored the data -
<Locations>
<location>
<place>Newyork</place>
<dt>01-Dec-2011</dt>
</location>
<location>
<place>Berlin</place>
<dt>02-Dec-2011</dt>
</location>
<location>
<place>Tokyo</place>
<dt>04-Dec-2011</dt>
</location>
</Location>
我想要实现的是-
如果访问被重新安排，我想替换 标记日期值.例如-如果柏林的访问日期更改，存储在 <dt> 标签中，那么如何编辑/替换相同的在 XML 文件中使用 XSLT ..?提前致谢 - 约翰
I want to replace the <dt> tags date value, if the visit is re-scheduled. For example-
If the visit date for Berlin is changed, stored in <dt> tags, then how to edit/replace the same
in the XML file using XSLT..? Thanks in advance - John
推荐答案
此转换展示了如何使用全局参数(此处使用内联元素建模)来指定(可能是多个)更新:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:my="my:my" >
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<my:updates>
<update place="Berlin" dt="11-Dec-2011"/>
</my:updates>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match=
"location
[place = document('')/*/my:updates/update/@place]
/dt/text()
">
<xsl:value-of select=
"document('')/*/my:updates/update
[@place = current()/../../place]
/@dt
"/>
</xsl:template>
</xsl:stylesheet>
应用于提供的 XML 文档时(已更正以使其格式正确):
When applied on the provided XML document (corrected to make it well-formed):
<Locations>
<location>
<place>Newyork</place>
<dt>01-Dec-2011</dt>
</location>
<location>
<place>Berlin</place>
<dt>02-Dec-2011</dt>
</location>
<location>
<place>Tokyo</place>
<dt>04-Dec-2011</dt>
</location>
</Locations>
产生想要的、正确的结果:
<Locations>
<location>
<place>Newyork</place>
<dt>01-Dec-2011</dt>
</location>
<location>
<place>Berlin</place>
<dt>11-Dec-2011</dt>
</location>
<location>
<place>Tokyo</place>
<dt>04-Dec-2011</dt>
</location>
</Locations>
说明:
身份规则按原样"复制每个节点.
只有一个覆盖模板 -- 匹配任何 dt 的文本节点子节点，其 place 兄弟的字符串值具有相应的 my:updates/update 元素.在此模板中，我们输出此对应 my:updates/update 元素的 dt 属性的值.
There is just one overriding template -- matching the text-node child of any dt whose place sibling's string value has a corresponding my:updates/update element. In this template we output the value of the dt attribute of this corresponding my:updates/update element.
请注意:在现实世界的转换中，内联 my:updates 元素将更好地替换为外部全局参数.阅读您的 XSLT 处理器的文档，了解如何将外部参数传递给转换——这取决于实现.
Do note: In a realworld transformation the inline my:updates element will be better replaced by an external, global parameter. Read your XSLT processor's documentation how to pass an external parameter to the transformation -- this is implementation-dependent.
更新:由于 OP 发现很难将此解决方案转换为使用全局、外部传递的 xsl:param 的解决方案，这里是转换后的解决方案:
UPDATE: As the OP has found it difficult to convert this solution to one using global, externally passed xsl:param, here is this converted solution:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ext="http://exslt.org/common">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:param name="pUpdates">
<update place="Berlin" dt="11-Dec-2011"/>
</xsl:param>
<xsl:variable name="vUpdates" select=
"ext:node-set($pUpdates)/*"/>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match="dt/text()">
<xsl:choose>
<xsl:when test="../../place=$vUpdates/@place">
<xsl:value-of select=
"$vUpdates[@place = current()/../../place]/@dt"/>
</xsl:when>
<xsl:otherwise>
<xsl:value-of select="."/>
</xsl:otherwise>
</xsl:choose>
</xsl:template>
</xsl:stylesheet>
当这个转换应用于同一个 XML 文档(上图)时，会产生同样正确和想要的结果:
<Locations>
<location>
<place>Newyork</place>
<dt>01-Dec-2011</dt>
</location>
<location>
<place>Berlin</place>
<dt>11-Dec-2011</dt>
</location>
<location>
<place>Tokyo</place>
<dt>04-Dec-2011</dt>
</location>
</Locations>
注意:在这个解决方案中，xsl:param 的值仍然被硬编码，这是我们使用 ext:node-set 的唯一原因() 扩展函数.如果参数确实是从外部传递过来的，那么这个RTF到正则树的转换是没有必要的，直接引用参数即可.
Do note: In this solution the xsl:param still has its value hardcoded and this is the only reason we are using the ext:node-set() extension function. If the parameter is really passed from outside, then this convertion from RTF to a regular tree isn't necessary and the parameter should be referenced directly.
此外，在 XSLT 1.0 中，我们必须更不精确地匹配，并在模板主体内使用比较(xsl:choose).这是因为在 XSLT 1.0 中不允许在匹配模式中引用变量/参数.
Also, in XSLT 1.0 we have to match more inexactly and to use comparisons (the xsl:choose) inside the body of the template. This is so because in XSLT 1.0 it isn't allowed to reference variables/parameters inside the match-pattern.
在 XSLT 2.0 中，这个限制已经消除，所以我们可以有一个更简单的转换:
In XSLT 2.0 this limitation has been eliminated, so we can just have a much simpler transformation:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output omit-xml-declaration="yes" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:param name="pUpdates">
<update place="Berlin" dt="11-Dec-2011"/>
</xsl:param>
<xsl:template match="node()|@*">
<xsl:copy>
<xsl:apply-templates select="node()|@*"/>
</xsl:copy>
</xsl:template>
<xsl:template match=
"location[place=$pUpdates/*/@place]/dt/text()">
<xsl:value-of select=
"$pUpdates/*[@place = current()/../../place]/@dt"/>
</xsl:template>
</xsl:stylesheet>
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