从没有任何API的XML文件生成java文件 [英] Generate java file from XML file without any API

问题描述

我如何生成Java文件，意味着生成类的方法属性的名称，而不使用任何来自XML的API

我的XML文件：源

 < class name =person> 
 
< Attribut type =int> Age< / Attribut> 
< Attribut type =String> Name< / Attribut> 
< / class> 
  

到java文件：

  public class person {
 int age; 
字符串名称; 
} 
  

您的帮助非常赞赏

谢谢

解决方案

如果您还想自己做。您可能会看到这个简化的例子。

首先警告：

• 它缺少适当的异常处理


• 它只适用于您提出的XML示例

示例代码

  import java.io.FileInputStream; 
 import java.io.IOException; 
 import javax.xml.stream.FactoryConfigurationError; 
 import javax.xml.stream.XMLInputFactory; 
 import javax.xml.stream.XMLStreamConstants; 
 import javax.xml.stream.XMLStreamException; 
 import javax.xml.stream.XMLStreamReader; 
 
 public class XMLStreamReaderDemo {
 
 public static void main（String [] args）throws Exception {
 String xmlFileName =source.xml; 
 StringBuilder javaSource = transform（xmlFileName）; 
 System.out.println（javaSource）; 
} 
 
 static StringBuilder transform（String xmlFileName）throws 
 FactoryConfigurationError，IOException，XMLStreamException {
 XMLInputFactory factory = XMLInputFactory.newInstance（）; 
 XMLStreamReader parser = null; 
 StringBuilder source = new StringBuilder（）; 
 try（FileInputStream inputStream = new FileInputStream（xmlFileName））{
 parser = factory.createXMLStreamReader（inputStream）; 
 while（parser.hasNext（））{
 switch（parser.getEventType（））{
 case XMLStreamConstants.START_ELEMENT：
 processStartElement（parser，source）; 
 break; 
 case XMLStreamConstants.CHARACTERS：
 processCharacters（parser，source）; 
 break; 
 
 case XMLStreamConstants.END_ELEMENT：
 processEndElement（parser，source）; 
 break; 
默认值：
 break; 
} 
 parser.next（）; 
} 
} finally {
 if（parser！= null）{
 parser.close（）; 
} 
} 
返回源; 
} 
 
 static void processEndElement（XMLStreamReader reader，StringBuilder sb）{
 String element = reader.getLocalName（）; 
 if（class.equals（element））{
 sb.append（}）; 
} else if（Attribut.equals（element））{
 sb.append（; \\\
）; 
} 
} 
 
 static void processCharacters（XMLStreamReader reader，StringBuilder sb）{
 if（！reader.isWhiteSpace（））{
 sb.append （）.append（reader.getText（））; 
} 
} 
 
 static void processStartElement（XMLStreamReader reader，StringBuilder sb）{
 String element = reader.getLocalName（）; 
 if（class.equals（element））{
 sb.append（public class）
 .append（reader.getAttributeValue（0））
 .append {\\\
）; 
} else if（Attribut.equals（element））{
 sb.append（）
 .append（reader.getAttributeValue（0））; 
} 
} 
} 
  

假设 source.xml 包含

 < class name =person> 
< Attribut type =int> Age< / Attribut> 
< Attribut type =String> Name< / Attribut> 
< / class> 
  

代码打印

  public class person {
 int Age; 
字符串名称; 
} 
  

剩下的唯一的事情：实现所有缺少的部分。如果这仍然包含太多XML API ...好...写你自己的解析器。 ; - ）

How can I generate Java file, means generate name of classes methods attributes, without using any API from XML

my XML file : Source

<class name="person">

<Attribut type="int">Age</Attribut>
<Attribut type="String">Name</Attribut>
</class>

to java file:

 public class person {
    int age;
    String Name;
 }

Your help is very appreciated

Thank you

解决方案

If you still want to do it your own. You might have a look on this simplified example.

Warnings first:

• it lacks proper Exception handling
• it is build only to work with your proposed XML example

The sample code

import java.io.FileInputStream;
import java.io.IOException;
import javax.xml.stream.FactoryConfigurationError;
import javax.xml.stream.XMLInputFactory;
import javax.xml.stream.XMLStreamConstants;
import javax.xml.stream.XMLStreamException;
import javax.xml.stream.XMLStreamReader;

public class XMLStreamReaderDemo {

    public static void main(String[] args) throws Exception {
        String xmlFileName = "source.xml";
        StringBuilder javaSource = transform(xmlFileName);
        System.out.println(javaSource);
    }

    static StringBuilder transform(String xmlFileName) throws
            FactoryConfigurationError, IOException, XMLStreamException {
        XMLInputFactory factory = XMLInputFactory.newInstance();
        XMLStreamReader parser = null;
        StringBuilder source = new StringBuilder();
        try (FileInputStream inputStream = new FileInputStream(xmlFileName)) {
            parser = factory.createXMLStreamReader(inputStream);
            while (parser.hasNext()) {
                switch (parser.getEventType()) {
                    case XMLStreamConstants.START_ELEMENT:
                        processStartElement(parser, source);
                        break;
                    case XMLStreamConstants.CHARACTERS:
                        processCharacters(parser, source);
                        break;

                    case XMLStreamConstants.END_ELEMENT:
                        processEndElement(parser, source);
                        break;
                    default:
                        break;
                }
                parser.next();
            }
        } finally {
            if (parser != null) {
                parser.close();
            }
        }
        return source;
    }

    static void processEndElement(XMLStreamReader reader, StringBuilder sb) {
        String element = reader.getLocalName();
        if ("class".equals(element)) {
            sb.append("}");
        } else if ("Attribut".equals(element)) {
            sb.append(";\n");
        }
    }

    static void processCharacters(XMLStreamReader reader, StringBuilder sb) {
        if (!reader.isWhiteSpace()) {
            sb.append(" ").append(reader.getText());
        }
    }

    static void processStartElement(XMLStreamReader reader, StringBuilder sb) {
        String element = reader.getLocalName();
        if ("class".equals(element)) {
            sb.append("public class ")
                    .append(reader.getAttributeValue(0))
                    .append(" {\n");
        } else if ("Attribut".equals(element)) {
            sb.append("    ")
                    .append(reader.getAttributeValue(0));
        }
    }
}

Assuming source.xml contains

<class name="person">
<Attribut type="int">Age</Attribut>
<Attribut type="String">Name</Attribut>
</class>

the code prints

public class person {
    int Age;
    String Name;
}

The "only" thing left for you to do: implement all the missing parts. If this still contains "too much" XML API ... well ... write you own parser. ;-)

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