java - 尝试从 .xsd 文件创建 Java 类时出错

Question

我有两个 XML，我想将它们转换为 java 类，所以我转到 this我可以在其中生成 .xsd 文件的页面。我试图在 Eclipse 中创建两个 Java 类，但遇到了相同的错误:

parsing a schema...
[ERROR] Content is not allowed in prolog.
  line 1 of file:/(...)/queryClass.xsd

Failed to parse a schema.

这些是我的 .xsd 文件:

第一:

<?xml version="1.0" encoding="utf-16"?>
<xsd:schema attributeFormDefault="unqualified" elementFormDefault="qualified" version="1.0" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <xsd:element name="dao" type="daoType" />
  <xsd:complexType name="daoType">
    <xsd:sequence>
      <xsd:element name="query" type="queryType" />
    </xsd:sequence>
  </xsd:complexType>
  <xsd:complexType name="queryType">
    <xsd:sequence>
      <xsd:element name="arguments" type="argumentsType" />
      <xsd:element name="statement" type="xsd:string" />
    </xsd:sequence>
    <xsd:attribute name="name" type="xsd:string" />
  </xsd:complexType>
  <xsd:complexType name="argumentsType">
    <xsd:sequence>
      <xsd:element name="argument" type="argumentType" />
    </xsd:sequence>
  </xsd:complexType>
  <xsd:complexType name="argumentType">
    <xsd:attribute name="name" type="xsd:string" />
  </xsd:complexType>
</xsd:schema>

第二:

<?xml version="1.0" encoding="utf-16"?>
<xsd:schema attributeFormDefault="unqualified" elementFormDefault="qualified" version="1.0" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <xsd:element name="properties" type="propertiesType" />
  <xsd:complexType name="propertiesType">
    <xsd:sequence>
      <xsd:element maxOccurs="unbounded" name="entry" type="entryType" />
    </xsd:sequence>
    <xsd:attribute name="version" type="xsd:decimal" />
  </xsd:complexType>
  <xsd:complexType name="entryType">
    <xsd:attribute name="key" type="xsd:string" />
  </xsd:complexType>
</xsd:schema>

Answer 1

事实证明，XSD 文件是错误的。我用过another XML-XSD converter而且效果非常好。