Question

我试图只从我的XML Author属性输出，等于 McGraw-Hill ，我是XSLT的新手我想知道我需要改变什么才能让它工作如预期的那样。

?xml version="1.0" encoding="UTF-8"?>

<xsl:stylesheet version="2.0" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
    <xsl:output method="html"/>
    <xsl:template match="/">
        <html>
            <head>
                <title>Book's list</title>
                    <link   type="text/css"  rel="stylesheet"  href="style.css"/>
            </head>
            <body>
                <xsl:apply-templates select="bookstore/book">
                <xsl:sort select="year" data-type="number" order="descending"/>
                <xsl:for-each select="bookstore/book/authors">
                    <xsl:if test="author = 'McGraw-Hill'">
                    </xsl:if>
                </xsl:for-each>
                </xsl:apply-templates>      
            </body>
        </html>
    </xsl:template>
    <xsl:template match="bookstore">
        <xsl:apply-templates/>
    </xsl:template>
    <xsl:template match="book">

        <div class="book">
            <br/>
            <strong> <xsl:value-of select="title" /> </strong>
            <span> Price: $ <xsl:value-of select="price" /> </span>
        </div>

        <div class="author">
            Author(s):
            <br/>
            <xsl:for-each select="authors/author" >
                <xsl:value-of select="."/>
                <br/>
            </xsl:for-each>
        </div>
    </xsl:template>
</xsl:stylesheet>

XML本身不包含作者，我希望它不输出任何内容，但它会在降序发布日期列出所有书籍，如果它不包含作者，则不希望包含书籍。< / p>

<?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type="text/xsl" href="Portfolio4.xsl"?>
<bookstore>

  <book category="cooking">
    <title lang="en">Everyday Italian</title>
    <authors>   
        <author>Giada De Laurentiis</author>
    </authors>
    <year>2005</year>
    <price>30.00</price>
  </book>

  <book category="children">
    <title lang="en">Harry Potter</title>
    <authors>
        <author>J K. Rowling</author>
    </authors>
    <year>2005</year>
    <price>29.99</price>
  </book>

  <book category="web">
    <title lang="en">XQuery Kick Start</title>
    <authors>
        <author>James McGovern</author>
        <author>Per Bothner</author>
        <author>Kurt Cagle</author>
        <author>James Linn</author>
        <author>Vaidyanathan Nagarajan</author>
    </authors>
    <year>2003</year>
    <price>49.99</price>
  </book>

  <book category="web" cover="paperback">
    <title lang="en">Learning XML</title>
    <authors>
        <author>Erik T. Ray</author>
    </authors>
    <year>2003</year>
    <price>39.95</price>
  </book>

  <book category= "engineering" cover="hardcover">
    <title lang="en">Network Fundamentals, CCNA Exploration Companion Guide</title>
    <authors>   
        <author>Mark Dye</author>
        <author>Rick McDonald</author>
        <author>Antoon Rufi</author>
    </authors>
    <year>2007</year>
    <price>7.25</price>
  </book>

  <book category= "engineering" cover="hardcover">
    <title lang="en">Modern Operating Systems</title>
    <authors>
        <author>Andrew S. Tanenbaum</author>
        <author>Herbert Bos</author>
    </authors>
    <year>2001</year>
    <price>14.94</price>
  </book>

 <book category= "engineering" cover="hardcover">
    <title lang="en">The Art of Computer Programming, Volume 1: Fundamental Algorithms</title>
    <authors>   
        <author>Donald Ervin Knuth</author>
    </authors>
    <year>1997</year>
    <price>75.32</price>
  </book>
</bookstore>

编辑：添加了XML，所需的输出将是包含我问题中所述作者的任何书籍将显示在输出中。

Answer 1

你的问题出现在这段代码中

appsettings.json

主要问题是你无法在<xsl:apply-templates select="bookstore/book"> <xsl:sort select="year" data-type="number" order="descending"/> <xsl:for-each select="bookstore/book/authors"> <xsl:if test="author = 'McGraw-Hill'"> </xsl:if> </xsl:for-each> </xsl:apply-templates>中嵌套xsl:for-each。

您应该将其简化为此

xsl:apply-templates

（当然，这不会在您的示例XML中返回任何结果，因为它没有任何由他们创作的书籍。）

试试这个XSLT

<xsl:apply-templates select="bookstore/book[authors/author='McGraw-Hill']">
    <xsl:sort select="year" data-type="number" order="descending"/>
</xsl:apply-templates>

需要我的XSLT仅从XML输出特定作者（如果可用）

1 个答案: