2

我只需要编写以下愚蠢的课程以避免发疯：

import java.io.OutputStream;

import org.w3c.dom.Document;
import org.w3c.dom.bootstrap.DOMImplementationRegistry;
import org.w3c.dom.ls.DOMImplementationLS;
import org.w3c.dom.ls.LSOutput;
import org.w3c.dom.ls.LSSerializer;

public final class XMLSerializer {
    public static final void writeDocument(Document input, OutputStream output) {
        try {
            DOMImplementationLS ls = 
                (DOMImplementationLS) DOMImplementationRegistry
                .newInstance().getDOMImplementation("LS");
            LSSerializer ser = ls.createLSSerializer();
            LSOutput out = ls.createLSOutput();
            out.setByteStream(output);
            ser.write(input, out);
        } catch (Exception e) { // DIAF Java
            throw new RuntimeException(e);
        }
    }
}

这种便捷方法是否已经存在于 Java 或通用库中？它似乎冗长可笑，甚至是所有异常都折叠在“catch (Exception e)”下的版本。

4

2 回答 2

0

尝试这个：

DOMSource domSource = new DOMSource(input);
StreamResult resultStream = new StreamResult(output);
TransformerFactory transformFactory = TransformerFactory.newInstance();
try {
    serializer.transform(domSource, resultStream);
} catch (javax.xml.transform.TransformerException e) {
}
于 2009-06-22T02:12:43.070 回答
0

尝试dom4j。它是一个优秀的 XML 库。

于 2009-06-22T04:21:48.487 回答