Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
751 views
in Technique[技术] by (71.8m points)

xslt - Sort XML nodes in alphabetical order using XSL

I am trying to figure out how to sort the XML list of employees alphabetically by lastname using XSL. Right now it just displays the XML information in the same order as it is in the XML. I don't think I fully understand how to use the <xsl:sort> function as I am new to XSL. I also tried putting order-by="+ Lastname" in with <xsl:for-each> and I couldn't get that to work either.

Heres my xml:

<?xml version="1.0"?>
<?xml-stylesheet type="text/xsl" href="Company1.xsl"?>
<Company>
    <Employee>
        <Firstname>John</Firstname>
        <Lastname>Smith</Lastname>
        <ssn>635-35-7463</ssn>
        <doh>February 3, 2011</doh>
        <Age>34</Age>
    </Employee>
    <Employee>
        <Firstname>Brad</Firstname>
        <Lastname>Roberts</Lastname>
        <ssn>789-65-4568</ssn>
        <doh>February 13, 2012</doh>
        <Age>25</Age>
    </Employee>
    <Employee>
        <Firstname>Karen</Firstname>
        <Lastname>Smith</Lastname>
        <ssn>369-12-7415</ssn>
        <doh>March 24, 2011</doh>
        <Age>28</Age>
    </Employee>
    <Employee>
        <Firstname>Eli</Firstname>
        <Lastname>Smith</Lastname>
        <ssn>489-32-8525</ssn>
        <doh>September 14, 2010</doh>
        <Age>38</Age>
    </Employee>
    <Employee>
        <Firstname>Bill</Firstname>
        <Lastname>Joel</Lastname>
        <ssn>689-67-7634</ssn>
        <doh>February 29, 2012</doh>
        <Age>24</Age>
    </Employee>
    <Employee>
        <Firstname>Kelly</Firstname>
        <Lastname>Greene</Lastname>
        <ssn>927-82-6873</ssn>
        <doh>December 3, 2010</doh>
        <Age>34</Age>
    </Employee>
</Company>

And heres my XSL:

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

    <xsl:template match="/">
    <HTML>
    <HEAD>
    <TITLE>Company Employees</TITLE> 
    </HEAD>
    <BODY>
    <H2>Company Employees</H2> 

    <xsl:for-each select="Company/Employee">
        <xsl:sort select="Employee/Lastname" data-type="text" order="ascending"/>
        <xsl:sort select="Employee/Firstname" data-type="text" order="ascending"/>

        <SPAN STYLE="font-weight:bold">FirstName: </SPAN>
            <xsl:value-of select="Lastname" />
        <BR />
        <SPAN STYLE="font-weight:bold">LastName: </SPAN>
            <xsl:value-of select="Firstname" />
        <BR /> 
        <SPAN STYLE="font-weight:bold">SSN: </SPAN> 
        <xsl:value-of select="ssn" /> 
        <BR /> 
        <SPAN STYLE="font-weight:bold">Date of Hire: </SPAN> 
        <xsl:value-of select="doh" /> 
        <BR /> 
        <SPAN STYLE="font-weight:bold">Age: </SPAN> 
        <xsl:value-of select="Age" /> 
        <P/>
    </xsl:for-each>
    </BODY>
    </HTML>
    </xsl:template>
</xsl:stylesheet>
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

At a glance,

    <xsl:sort select="Employee/Lastname" data-type="text" order="ascending"/>
    <xsl:sort select="Employee/Firstname" data-type="text" order="ascending"/>

should be

    <xsl:sort select="Lastname" data-type="text" order="ascending"/>
    <xsl:sort select="Firstname" data-type="text" order="ascending"/>

for-each sets the context node for the select, so the expression is evaluated against the Employee nodes.

Also, text and ascending are defaults, so you could just write

    <xsl:sort select="Lastname"/>
    <xsl:sort select="Firstname"/>

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...