温馨提示:本文翻译自stackoverflow.com,查看原文请点击:其他 - Error while trying to create Java Class out of .xsd file
java xsd

其他 - 尝试从.xsd文件创建Java类时出错

发布于 2020-04-22 18:46:55

我有两个个XML,我想将它们转换成Java类,所以我去了这个网页,我可以生成xsd文件。我想在Eclipse中使用它们制作两个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>

查看更多

提问者
CoralWombat
被浏览
34
CoralWombat 2020-02-07 00:16

原来,XSD文件是错误的。我使用了另一个XML-XSD转换器,它运行良好。