xml：如何在 .xml 文件中引用 .xsd 文件？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/3965357/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
xml : how to reference a .xsd file at .xml file?
提问by RedsDevils
I want to see xml file in browser as I define in .xsd file. Please check the following two files for me and point out what do I need to do. These two files are under same folder.
我想在浏览器中查看 .xsd 文件中定义的 xml 文件。请帮我检查以下两个文件，并指出我需要做什么。这两个文件在同一个文件夹下。
employee.xml
员工文件
<?xml version="1.0"?>
<employee xmlns="http://www.w3schools.com"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="employee.xsd">
<firstname>John</firstname>
<lastname>Smith</lastname>
</employee>
employee.xsd
员工.xsd
<xs:element name="employee">
<xs:complexType>
<xs:sequence>
<xs:element name="firstname" type="xs:string" fixed="red" />
<xs:element name="lastname" type="xs:string"/>
</xs:sequence>
</xs:complexType>
</xs:element>
回答by Oleg
You made two errors: one in the schema file and another in the syntax of the value of the xsi:schemaLocationattribute of the XML file.
您犯了两个错误：一个在架构文件中，另一个在xsi:schemaLocationXML 文件的属性值的语法中。
The main error is that your employee.xsd file is only a fragment of the XML Schema. You should complete the contain of the employee.xsd. For example,
主要错误是您的employee.xsd 文件只是XML Schema 的一个片段。您应该完成对employee.xsd 的包含。例如，
<?xml version="1.0" encoding="utf-8"?>
<xs:schema targetNamespace="http://www.w3schools.com/RedsDevils"
elementFormDefault="qualified"
xmlns="http://www.w3schools.com/RedsDevils employee.xsd"
xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name="employee">
<xs:complexType>
<xs:sequence>
<xs:element name="firstname" type="xs:string" fixed="red" />
<xs:element name="lastname" type="xs:string"/>
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>
and employee.xml:
和员工.xml：
<?xml version="1.0" encoding="utf-8"?>
<employee xmlns="http://www.w3schools.com/RedsDevils"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.w3schools.com/RedsDevils employee.xsd">
<firstname>John</firstname>
<lastname>Smith</lastname>
</employee>
Because you define default namespace in the XML file, the schema location attribule xsi:schemaLocationmust consist from the the namespace and the path to the schema devided with the blank. I changed the namespace name so that it will be a little more unique: "http://www.w3schools.com/RedsDevils"instead of "http://www.w3schools.com".
因为您在 XML 文件中定义了默认命名空间，所以架构位置属性xsi:schemaLocation必须由命名空间和以空格分隔的架构路径组成。我更改了命名空间名称，使其更加独特："http://www.w3schools.com/RedsDevils"而不是"http://www.w3schools.com".
At the end I can add that the XML file employee.xml not corresponds to the schema employee.xsd because the element <firstname>John</firstname>has the value other as red, but probably exactly this you wanted to test.
最后，我可以补充一点，XML 文件employee.xml 与架构employee.xsd 不对应，因为该元素<firstname>John</firstname>的值为 other as red，但可能正是您想要测试的。
