使用JAXB解组XML而不会转义字符
想象以下情况:我们从某个外部工具收到一个xml文件。最近,在此xml中,节点名称中或它们的richcontent标记中可能有一些转义的字符,如以下示例所示(简化):
<map><node TEXT="Project">
<node TEXT="ää">
<richcontent TYPE="NOTE"><html>
  <head>
  </head>
  <body>
    <p>
      I am a Note for Node ää!
    </p>
  </body>
</html>
</richcontent>
</node>
</node>
</map>
在使用JAXB解组文件后,那些逃脱的字符将不被转义。不幸的是,我需要他们保持原样,这意味着逃脱了。有什么方法可以避免在解组时将这些字符转义?
使用JAXB甚至有可能实现这一目标,还是我们甚至必须考虑更改为其他xml阅读器API?
预先谢谢你,艾梅娜
回答:
你只需要更换&#的&#,因此呼叫
unmarshaller.unmarshal(new AmpersandingStream(new FileInputStream(...)));和
import java.io.IOException;import java.io.InputStream;
/**
* Replaces numerical entities with their notation as text.
*/
public class AmpersandingStream extends InputStream {
    private InputStream in;
    private boolean justReadAmpersand;
    private String lookAhead = "";
    public AmpersandingStream(InputStream in) {
        this.in = in;
    }
    @Override
    public int read() throws IOException {
        if (!lookAhead.isEmpty()) {
            int c = lookAhead.codePointAt(0);
            lookAhead = lookAhead.substring(Character.charCount(c));
            return c;
        }
        int c = in.read();
        if (c == (int)'#' && justReadAmpersand) {
            c = (int)'a';
            lookAhead = "mp;#";
        }
        justReadAmpersand = c == (int)'&';
        return c;
    }
    @Override
    public int available() throws IOException {
        return in.available();
    }
    @Override
    public void close() throws IOException {
        in.close();
    }
    @Override
    public synchronized void mark(int readlimit) {
        in.mark(readlimit);
    }
    @Override
    public boolean markSupported() {
        return in.markSupported();
    }
    @Override
    public int read(byte[] b) throws IOException {
        return in.read(b);
    }
    @Override
    public int read(byte[] b, int off, int len) throws IOException {
        return in.read(b, off, len);
    }
    @Override
    public synchronized void reset() throws IOException {
        in.reset();
    }
    @Override
    public long skip(long n) throws IOException {
        return in.skip(n);
    }
}
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