通过 XPath 中的 concat() 换行 (\n)？

Question

所以我的系统中有一个员工的 XML 文档：

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<couriersystem title="System"
    xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:noNamespaceSchemaLocation="schema.xsd">
    <!-- snip -->
    <employees>
        <employee eid="1">
            <nin>AZ123518D</nin>
            <firstname>Peter</firstname>
            <lastname>Smith</lastname>
            <gender>Male</gender>
            <dob>1994-02-11</dob>
            <email>ps11@gmail.com</email>
            <address>
                119, London Street, Nidrie, F57 8NE
            </address>
            <tel>07005748900</tel>
            <salary>30526</salary>
            <empbranch bid="1" />
            <supervisor sid="1" />
        </employee>
        <employee eid="2">
            <nin>CN174869F</nin>
            <firstname>Jennifer</firstname>
            <lastname>Black</lastname>
            <gender>Male</gender>
            <dob>1984-12-24</dob>
            <email>jb21@gmail.com</email>
            <address>
                161, South Road, Nidrie, W79 8WG
            </address>
            <tel>07555111222</tel>
            <salary>40576</salary>
            <empbranch bid="2" />
            <supervisor sid="1" />
        </employee>
        <employee eid="3">
            <nin>ET127654M</nin>
            <firstname>Aaron</firstname>
            <lastname>Jones</lastname>
            <gender>Male</gender>
            <dob>1968-03-15</dob>
            <email>aj31@gmail.com</email>
            <address>
                66, High Road, Yoker, Q47 4SR
            </address>
            <tel>07856471267</tel>
            <salary>30526</salary>
            <empbranch bid="3" />
            <supervisor sid="1" />
        </employee>
        <employee eid="4">
            <nin>GC765238A</nin>
            <firstname>Alistair</firstname>
            <lastname>Smith</lastname>
            <gender>Male</gender>
            <dob>1976-11-26</dob>
            <email>as11@gmail.com</email>
            <address>
                109, West Plaza, Clydebank, G55 8RC
            </address>
            <tel>07000123123</tel>
            <salary>25400</salary>
            <empbranch bid="4" />
            <supervisor sid="1" />
        </employee>
        <employee eid="5">
            <nin>HP146854D</nin>
            <firstname>Emma</firstname>
            <lastname>Reynolds</lastname>
            <gender>Male</gender>
            <dob>1995-05-05</dob>
            <email>er11@yahoo.com</email>
            <address>
                57, Scott Street, Aberdeen, O75 2KS
            </address>
            <tel>07625361536</tel>
            <salary>25400</salary>
            <empbranch bid="5" />
            <supervisor sid="7" />
        </employee>
        <!-- snip -->
    </employees>
    <!-- snip -->
</couriersystem>

我有以下 XPath 可以由他们的主管检索所有员工的详细信息：

//employee[supervisor/@sid='1']/concat('Name: ', concat(firstname/text(), ' ', lastname/text(), '\nGender: ', gender/text(), '\nD.O.B: ', dob/text()))

但它只显示其余员工中的一名员工，\n 字符也会显示。

我该如何解决这个问题？

Answer 1

要在 XPath 中将 \n 解释为 concat()，请使用 codepoints-to-string(10) [credit: @madcrazydrumma]:

//employee[supervisor/@sid='1']/concat('Name: ', firstname, ' ', lastname, 
                                       codepoints-to-string(10),
                                       'Gender: ', gender, 
                                       codepoints-to-string(10),
                                       'D.O.B: ', dob)

然后返回结果如下：

Name: Peter Smith
Gender: Male
D.O.B: 1994-02-11
Name: Jennifer Black
Gender: Male
D.O.B: 1984-12-24
Name: Aaron Jones
Gender: Male
D.O.B: 1968-03-15
Name: Alistair Smith
Gender: Male
D.O.B: 1976-11-26

---

小提示：您的基本 XPath 2.0 表达式结果很好。正如我们在 cmets 中发现的那样，由于站点 xpathtester.com/xpath 的限制，结果不正确。使用 videlibri.sourceforge.net/cgi-bin/xidelcgi 代替在线 XPath 2.0 测试。

Answer 2

好吧，我用 XSLT 测试了你的 XPath 表达式，输出包含你想要的 4 个条目：

我的 XSLT 是：

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="/couriersystem">
    <xsl:copy-of select="//employee[supervisor/@sid='1']/concat('Name: ', concat(firstname/text(), ' ', lastname/text(), '\nGender: ', gender/text(), '\nD.O.B: ', dob/text()))" />
  </xsl:template>
</xsl:stylesheet>

结果都在一行文本中：

<?xml version="1.0" encoding="UTF-8"?>Name: Peter Smith\nGender: Male\nD.O.B: 1994-02-11 Name: Jennifer Black\nGender: Male\nD.O.B: 1984-12-24 Name: Aaron Jones\nGender: Male\nD.O.B: 1968-03-15 Name: Alistair Smith\nGender: Male\nD.O.B: 1976-11-26

如果你想让输出的格式比上面的多一点，试试这个在适当的地方插入space字符的表达式：

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:template match="/couriersystem">
    <xsl:value-of select="'&#10;'" />
    <xsl:for-each select="//employee[supervisor/@sid='1']">
      <xsl:value-of select="concat('Name: ', firstname/text(), ' ', lastname/text(), '&#10;','Gender: ', gender/text(), '&#10;','D.O.B: ', dob/text(),'&#10;')" />
    </xsl:for-each>
  </xsl:template>
</xsl:stylesheet>

这样的结果是：

<?xml version="1.0" encoding="UTF-8"?>
Name: Peter Smith
Gender: Male
D.O.B: 1994-02-11
Name: Jennifer Black
Gender: Male
D.O.B: 1984-12-24
Name: Aaron Jones
Gender: Male
D.O.B: 1968-03-15
Name: Alistair Smith
Gender: Male
D.O.B: 1976-11-26

P.S.：“詹妮弗·布莱克”是变性人吗？因为她有一个女性名字和一个男性性别;-)