Xml 在XPath中通过concat()换行(\n)?
因此,我的系统中有一个员工的XML文档:Xml 在XPath中通过concat()换行(\n)?,xml,xpath,newline,Xml,Xpath,Newline,因此,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 -->
<?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>
但它只显示其余员工中的一名,同时也显示\n
字符。
如何解决此问题?若要在XPath中的
concat()
中解释\n
,请使用codepoints到字符串(10)
[credit::
然后将返回如下结果:
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
次要说明:您的基本XPath 2.0表达式没有问题。正如我们在评论中发现的,由于网站的限制,结果不正确。请改为用于在线XPath 2.0测试。好的,我用XSLT测试了您的XPath表达式,输出包含您想要的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>
其结果是:
<?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
姓名:彼得·史密斯
性别:男
D.O.B:1994-02-11
姓名:詹妮弗·布莱克
性别:男
D.O.B:1984-12-24
姓名:亚伦·琼斯
性别:男
D.O.B:1968-03-15
姓名:阿利斯泰史密斯
性别:男
D.O.B:1976-11-26
旁白:“Jennifer Black”是变性人吗?因为她有一个女性名字和一个男性;-)但它只返回一个结果,这是最奇怪的事情!你在使用什么XPath 2.0库?我看到该站点的XPath 2.0实现存在这样的问题。请尝试该站点:@kjhuges解决了这个问题!但是我如何使用新的XPath 2.0库呢行?我使用
代码点来表示字符串(10)
在XPath 2.0中添加新行。如果您想向其他人展示,您可以更新您的代码,谢谢您的新站点!哈哈,我的朋友输入了数据,所以可能吧!;)我没有为此使用XSL,所以很遗憾,这并不能解决问题。我的最后一句话只是一个小笑话g。然而,我的XPath表达式基本上与acc相同接受答案,只需将和#10;
替换为codepoints to string(10)
,然后合并表达式和您的完成。
<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>
<?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
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/couriersystem">
<xsl:value-of select="' '" />
<xsl:for-each select="//employee[supervisor/@sid='1']">
<xsl:value-of select="concat('Name: ', firstname/text(), ' ', lastname/text(), ' ','Gender: ', gender/text(), ' ','D.O.B: ', dob/text(),' ')" />
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
<?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