我有以下xml
<R N="14" MIME="application/pdf">
<RK>7</RK>
<MT N="Abstract"
V="Lorem Ipsum is simply dummy text of the printing " />
<MT N="Abstract1"
V="and typesetting industry. Lorem Ipsum has been the industry's standard "/>
<MT N="Author" V="Bernard Shaw;" />
<MT N="Author1" V="Mark Twain" />
<MT N="Abstract2"
V="dummy text ever since the 1500s, when an unknown printer took a galley"/>
<LANG>en</LANG>
</R>
使用XSLT进行转换时,我需要连接Abstract和Author字段,并将其显示为
Abstract: Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley
Author: Bernard Shaw;Mark Twain
Abstract,Abstract1,Abstract2可以在xml中以任何顺序出现。
我正在尝试使用类似下面的东西,但坚持使用条件&amp;当Abstract,Abstract1没有以相同的顺序出现时串联串联
<xsl:template match="MT">
<xsl:if test="(some generic condition to display Title)">
<br/>
<span class="f">
<xsl:value-of select="@N"/>
</span>
</xsl:if>
<xsl:value-of select="@V"/>
</xsl:template>
感谢任何帮助。
答案 0 :(得分:2)
此转化:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/">
Abstract: <xsl:text/>
<xsl:for-each select="*/MT[starts-with(@N, 'Abstract')]">
<xsl:value-of select="@V"/>
</xsl:for-each>
Author: <xsl:text/>
<xsl:for-each select="*/MT[starts-with(@N, 'Author')]">
<xsl:value-of select="@V"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
应用于提供的XML文档时:
<R N="14" MIME="application/pdf">
<RK>7</RK>
<MT N="Abstract" V="Lorem Ipsum is simply dummy text of the printing " />
<MT N="Abstract1" V="and typesetting industry. Lorem Ipsum has been the industry's standard " />
<MT N="Author" V="Bernard Shaw;" />
<MT N="Author1" V="Mark Twain" />
<MT N="Abstract2" V="dummy text ever since the 1500s, when an unknown printer took a galley" />
<LANG>en</LANG>
</R>
产生想要的结果:
Abstract: Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley
Author: Bernard Shaw;Mark Twain
进一步重构:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/">
Abstract: <xsl:text/>
<xsl:call-template name="concatAttributes"/>
Author: <xsl:text/>
<xsl:call-template name="concatAttributes">
<xsl:with-param name="pKeyStartString" select="'Author'"/>
</xsl:call-template>
</xsl:template>
<xsl:template name="concatAttributes">
<xsl:param name="pKeyAttribName" select="'N'"/>
<xsl:param name="pKeyStartString" select="'Abstract'"/>
<xsl:param name="pValueAttribName" select="'V'"/>
<xsl:for-each select=
"*/MT[starts-with(@*[name()=$pKeyAttribName], $pKeyStartString)]">
<xsl:value-of select="@*[name()=$pValueAttribName]"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
第二次重构(OP请求):
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="text"/>
<xsl:strip-space elements="*"/>
<xsl:variable name="vAbstract">
<xsl:apply-templates mode="retrieve" select="//MT"/>
</xsl:variable>
<xsl:variable name="vAuthors">
<xsl:apply-templates mode="retrieve" select="//MT">
<xsl:with-param name="pKeyStartString" select="'Author'"/>
</xsl:apply-templates>
</xsl:variable>
<xsl:template match="/">
Abstract: <xsl:value-of select="$vAbstract"/>
Authors:: <xsl:value-of select="$vAuthors"/>
</xsl:template>
<xsl:template match="MT" mode="retrieve">
<xsl:param name="pKeyAttribName" select="'N'"/>
<xsl:param name="pKeyStartString" select="'Abstract'"/>
<xsl:param name="pValueAttribName" select="'V'"/>
<xsl:if test="starts-with(@*[name()=$pKeyAttribName], $pKeyStartString)">
<xsl:value-of select="@*[name()=$pValueAttribName]"/>
</xsl:if>
</xsl:template>
</xsl:stylesheet>
编辑:OP已请求“所有处理都应在匹配MT
的模板中完成”。虽然在简单的情况下这是可能的(参见@Alejandro的答案),但在匹配MT
的单个模板中进行所有处理会打开未知数的巨大空白。例如,可能需要以不同的方式处理其他MT
元素,在这种情况下,根本不会进行此类处理。
在更复杂的情况下(例如当元素和属性以任何顺序排列,但输出必须排序 - Abstract1,Abstract2,...,Abstract-N)时,必须明确指定排序,这必须在匹配MT
的模板之外。 因此,在一般情况下,只能在匹配MT
的模板中使用代码生成所需的输出。
我建议匹配MT
的单个模板处于命名模式,并且应该在“拉式”中使用 - 由调用者明确应用。
答案 1 :(得分:0)
<xsl:for-each select="MT[starts-with(@N, 'Abstract')]">
<xsl:if test="position() = 1">
<xsl:value-of select="@N"/>:
</xsl:if>
<xsl:value-of select="@V"/>
</xsl:for-each>
答案 2 :(得分:0)
此样式表:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:key name="kMtByAlphaN"
match="MT"
use="concat(generate-id(..),'+',translate(@N,'1234567890',''))"/>
<xsl:template match="MT[count(.|key('kMtByAlphaN',
concat(
generate-id(..),'+',
translate(@N,'1234567890','')
)
)[1]
) = 1
]">
<xsl:variable name="vName"
select="translate(@N,'1234567890','')"/>
<xsl:value-of select="concat($vName,': ')"/>
<xsl:apply-templates select="key('kMtByAlphaN',
concat(generate-id(..),'+',$vName)
)/@V"/>
<xsl:text>
</xsl:text>
</xsl:template>
<xsl:template match="text()"/>
</xsl:stylesheet>
输出:
Abstract: Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley
Author: Bernard Shaw;Mark Twain
注意:按@N字母内容和父级生成的ID进行分组,因为我希望有多个R
元素。