我需要一些xsl转换的帮助,我不知道如何开始,因为我是一个新手。
我有这个xml方案:
<?xml version="1.0" encoding="utf-8"?>
<GetUserCollectionFromSiteResponse xmlns="http://schemas.microsoft.com/sharepoint/soap/directory/">
<GetUserCollectionFromSiteResult>
<GetUserCollectionFromSite>
<Users>
<User ID="87" Sid="S-1-5-21-2025429265-1935655697-839522115-7617" Name="Falco Lannoo" LoginName="Domain\flannoo" Email="falco.lannoo@email.com" Notes="" IsSiteAdmin="False" IsDomainGroup="False" />
<User ID="31" Sid="S-1-5-21-2025429265-1935655697-839522115-2721" Name="John Smith" LoginName="Domain\jsmith" Email="john.smith@email.com" Notes="" IsSiteAdmin="False" IsDomainGroup="False" />
</Users>
</GetUserCollectionFromSite>
</GetUserCollectionFromSiteResult>
我想把它变成这个:
<ns0:userInfo xmlns:ns0="http://Sharepoint.userInfo">
<ID>218</ID>
<Name>Falco Lannoo</Name>
</ns0:userInfo>
所以我想选择loginname =“Domain \ flannoo”的节点。 任何人都可以帮助我进行这种转换,它必须在XSLT 1.0中
谢谢
答案 0 :(得分:1)
此样式表:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ns0="http://Sharepoint.userInfo"
xmlns:soap="http://schemas.microsoft.com/sharepoint/soap/directory/"
exclude-result-prefixes="soap">
<xsl:template match="soap:User[@LoginName='Domain\flannoo']">
<ns0:userInfo>
<xsl:apply-templates select="@*" />
</ns0:userInfo>
</xsl:template>
<xsl:template match="@*"/>
<xsl:template match="@ID|@Name">
<xsl:element name="{name()}">
<xsl:value-of select="." />
</xsl:element>
</xsl:template>
</xsl:stylesheet>
正确输入:
<GetUserCollectionFromSiteResponse xmlns="http://schemas.microsoft.com/sharepoint/soap/directory/">
<GetUserCollectionFromSiteResult>
<GetUserCollectionFromSite>
<Users>
<User ID="87" Sid="S-1-5-21-2025429265-1935655697-839522115-7617" Name="Falco Lannoo" LoginName="Domain\flannoo" Email="falco.lannoo@email.com" Notes="" IsSiteAdmin="False" IsDomainGroup="False" />
<User ID="31" Sid="S-1-5-21-2025429265-1935655697-839522115-2721" Name="John Smith" LoginName="Domain\jsmith" Email="john.smith@email.com" Notes="" IsSiteAdmin="False" IsDomainGroup="False" />
</Users>
</GetUserCollectionFromSite>
</GetUserCollectionFromSiteResult>
</GetUserCollectionFromSiteResponse>
输出:
<ns0:userInfo xmlns:ns0="http://Sharepoint.userInfo">
<ID>87</ID>
<Name>Falco Lannoo</Name>
</ns0:userInfo>
答案 1 :(得分:0)
以下是亚历杭德罗回答的替代方案。这主要是一个样式问题,但如果你必须将它集成在一个更复杂的样式表中,这可能是相关的。
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:ns0="http://Sharepoint.userInfo"
xmlns:soap="http://schemas.microsoft.com/sharepoint/soap/directory/"
exclude-result-prefixes="soap">
<xsl:template match="/">
<xsl:apply-templates select="//soap:User[@ID='87']"/>
</xsl:template>
<xsl:template match="soap:User">
<ns0:userInfo>
<ID><xsl:value-of select="@ID"/></ID>
<Name><xsl:value-of select="@Name"/></Name>
</ns0:userInfo>
</xsl:template>
</xsl:stylesheet>