使用XSD时xmllint验证错误"找不到DTD"

Question

我正在尝试使用xmllint来检查我的工作,同时开发一个基本的XSD即XML Schema架构.但它给了我一个错误

验证失败:未找到DTD.

我究竟做错了什么？

我的xmllint命令:

xmllint --noout --valid --schema simple.xsd lucas-basic.xml
lucas-basic.xml:5: validity error : Validation failed: no DTD found !
        >
        ^
lucas-basic.xml validates

测试XSD文件:

<?xml version = "1.0" encoding = "UTF-8"?>
<!--Generated by XML Authority. Conforms to w3c http://www.w3.org/2001/XMLSchema-->
<xsd:schema xmlns:xsd = "http://www.w3.org/2001/XMLSchema">
        <xsd:element name = "vehicles">
                <xsd:complexType>
                        <xsd:sequence>
                                <xsd:element name = "nickname" type = "xsd:string" maxOccurs = "unbounded"/>
                        </xsd:sequence>
                </xsd:complexType>
        </xsd:element>

测试XML文件:

<?xml version="1.0"?>
<vehicles
                xmlns:xsi = "http://www.w3.org/2001/XMLSchema-instance"
                xsi:noNamespaceSchemaLocation = "http://lucas.ucs.ed.ac.uk/xml-schema/xmlns/simple.xsd"
        >
        <nickname>Bog Hopper</nickname>
        <nickname>Wee Beastie</nickname>
        <nickname>Count Zero</nickname>
</vehicles>

xsi:noNamespaceSchemaLocation上的URL确实返回上面的XSD.此外,我下载了xsd文件并将其作为额外措施放入当前目录,但这似乎没有改变任何东西.

Answer 1

--valid 用于DTD,而不是XSD.

这对你有用:

xmllint --noout --schema http://lucas.ucs.ed.ac.uk/xml-schema/xmlns/simple.xsd lucas-basic.xml