XPath從XML文件中選擇不同的值

Question

我有以下XML文件

<people>
<person>
    <name>Diana B. Aust</name>
    <address>8325 Meadow Rd</address>
    <city>Dallas, TX</city>
</person>
<person>
    <name>Diana C. Aust</name>
    <address>8325 Meadow Rd</address>
    <city>Dallas, TX</city>
</person>
<person>
    <name>Acelia T. Peguero</name>
    <address>59 Terry Ave</address>
    <city>Amityville, NY</city>
</person>
<person>
    <name>Acelia U. Peguero</name>
    <address>58 Terry Ave</address>
    <city>Amityville, NY</city>
</person>
</people>

我想根據唯一地址+城市組合選擇所有人。 我將如何使用XPath查詢執行此操作？ 正確的查詢應返回節點＃1，＃3和＃4。

Answer 1

XPath 2.0解決方案：

/*/person[not(concat(city, address)=preceding-sibling::*/concat(city, address))]

XSLT 1.0 Muenchian方法解決方案（僅供參考）：

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output omit-xml-declaration="yes" indent="yes"/>
    <xsl:strip-space elements="*"/>
    <xsl:key name="byAddressAndCity" match="person" 
             use="concat(address, '|', city)"/>
    <xsl:template match="/">
        <xsl:copy-of
            select="*/person[generate-id()=generate-id(
                key('byAddressAndCity', concat(address, '|', city))[1])]"/>
    </xsl:template>
</xsl:stylesheet>