Java 如何从字符串中的 XML 加载 org.w3c.dom.Document？
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How do I load an org.w3c.dom.Document from XML in a string?
提问by Frank Krueger
I have a complete XML document in a string and would like a Documentobject. Google turns up all sorts of garbage. What is the simplest solution? (In Java 1.5)
我在一个字符串中有一个完整的 XML 文档，并且想要一个Document对象。谷歌会出现各种垃圾。最简单的解决方案是什么？（在 Java 1.5 中）
SolutionThanks to Matt McMinn, I have settled on this implementation. It has the right level of input flexibility and exception granularity for me. (It's good to know if the error came from malformed XML - SAXException- or just bad IO - IOException.)
解决方案感谢Matt McMinn，我已经确定了这个实现。它具有适合我的输入灵活性和异常粒度级别。（很高兴知道错误是否来自格式错误的 XML - SAXException- 或者只是错误的 IO - IOException。）
public static org.w3c.dom.Document loadXMLFrom(String xml)
throws org.xml.sax.SAXException, java.io.IOException {
return loadXMLFrom(new java.io.ByteArrayInputStream(xml.getBytes()));
}
public static org.w3c.dom.Document loadXMLFrom(java.io.InputStream is)
throws org.xml.sax.SAXException, java.io.IOException {
javax.xml.parsers.DocumentBuilderFactory factory =
javax.xml.parsers.DocumentBuilderFactory.newInstance();
factory.setNamespaceAware(true);
javax.xml.parsers.DocumentBuilder builder = null;
try {
builder = factory.newDocumentBuilder();
}
catch (javax.xml.parsers.ParserConfigurationException ex) {
}
org.w3c.dom.Document doc = builder.parse(is);
is.close();
return doc;
}
采纳答案by Matt McMinn
This works for me in Java 1.5 - I stripped out specific exceptions for readability.
这在 Java 1.5 中对我有用 - 为了可读性，我去掉了特定的异常。
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import org.w3c.dom.Document;
import java.io.ByteArrayInputStream;
public Document loadXMLFromString(String xml) throws Exception
{
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
factory.setNamespaceAware(true);
DocumentBuilder builder = factory.newDocumentBuilder();
return builder.parse(new ByteArrayInputStream(xml.getBytes()));
}
回答by erickson
Whoa there!
哇哦！
There's a potentially serious problem with this code, because it ignores the character encoding specified in the String(which is UTF-8 by default). When you call String.getBytes()the platform default encoding is used to encode Unicode characters to bytes. So, the parser may think it's getting UTF-8 data when in fact it's getting EBCDIC or something… not pretty!
此代码存在潜在的严重问题，因为它忽略了中指定的字符编码String（默认为 UTF-8）。当您调用String.getBytes()平台时，默认编码用于将 Unicode 字符编码为字节。因此，解析器可能认为它正在获取 UTF-8 数据，而实际上它正在获取 EBCDIC 或其他东西……不漂亮！
Instead, use the parse method that takes an InputSource, which can be constructed with a Reader, like this:
相反，使用带有 InputSource 的 parse 方法，它可以用 Reader 构造，如下所示：
import java.io.StringReader;
import org.xml.sax.InputSource;
…
return builder.parse(new InputSource(new StringReader(xml)));
It may not seem like a big deal, but ignorance of character encoding issues leads to insidious code rot akin to y2k.
这似乎没什么大不了的，但是对字符编码问题的无知会导致类似于 y2k 的阴险代码腐烂。
回答by shsteimer
Just had a similar problem, except i needed a NodeList and not a Document, here's what I came up with. It's mostly the same solution as before, augmented to get the root element down as a NodeList and using erickson's suggestion of using an InputSource instead for character encoding issues.
刚刚遇到了类似的问题，除了我需要一个 NodeList 而不是一个文档，这就是我想出的。它与以前的解决方案大致相同，增加了将根元素作为 NodeList 并使用 erickson 的建议，即使用 InputSource 而不是字符编码问题。
private String DOC_ROOT="root";
String xml=getXmlString();
Document xmlDoc=loadXMLFrom(xml);
Element template=xmlDoc.getDocumentElement();
NodeList nodes=xmlDoc.getElementsByTagName(DOC_ROOT);
public static Document loadXMLFrom(String xml) throws Exception {
InputSource is= new InputSource(new StringReader(xml));
DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
factory.setNamespaceAware(true);
DocumentBuilder builder = null;
builder = factory.newDocumentBuilder();
Document doc = builder.parse(is);
return doc;
}
回答by Xavier Dury
To manipulate XML in Java, I always tend to use the Transformer API:
为了在 Java 中操作 XML，我总是倾向于使用 Transformer API：
import javax.xml.transform.Source;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMResult;
import javax.xml.transform.stream.StreamSource;
public static Document loadXMLFrom(String xml) throws TransformerException {
Source source = new StreamSource(new StringReader(xml));
DOMResult result = new DOMResult();
TransformerFactory.newInstance().newTransformer().transform(source , result);
return (Document) result.getNode();
}
