尝试从 .xsd 文件创建 Java Class 时出错

Question

我有两个 XML，我想将它们转换成 java 类，所以我转到 this 页面，在那里我可以生成 .xsd 文件。我试图在 Eclipse 中制作两个 Java Class，但我遇到了同样的错误：

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：

第一个：

<?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>

第二个：

<?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

原来是XSD文件错了。我用了 another XML-XSD converter 并且效果很好。

尝试从 .xsd 文件创建 Java Class 时出错

Error while trying to create Java Class out of .xsd file

java

xsd