java 如何使用 JAXB2.0 禁用 DTD 获取
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How to disable DTD fetching using JAXB2.0
提问by Nick
I'm trying to use JAXB to unmashall some XML which I used xjc to create in the first place. I don't want to do any validation on the unmarshalling, but even though I have disabled the validation according to the JAXB documentation with u.setSchema(null);, but this hasn't prevented a FileNotFoundExceptionbeing thrown when it tries to run and can't find the schema.
我正在尝试使用 JAXB 来解散一些我首先使用 xjc 创建的 XML。我不想对解组进行任何验证，但即使我根据 JAXB 文档禁用了验证u.setSchema(null);，但这并没有阻止FileNotFoundException在它尝试运行并且找不到模式时抛出。
JAXBContext jc = JAXBContext.newInstance("blast");
Unmarshaller u = jc.createUnmarshaller();
u.setSchema(null);
return u.unmarshal(blast)
I've seen similar questions for disabling SAX parsing from validation by setting the apache property http://apache.org/xml/features/validation/schemato false, but I can't get the Unmarshaller to use my own sax parser.
我已经看到了致残SAX通过设置apache的财产验证解析类似的问题http://apache.org/xml/features/validation/schema来false，但我不能让着Unmarshaller用自己的SAX解析器。
采纳答案by bdoughan
Below is sample code that demonstrates how to get a JAXB (JSR-222)implementation to use your SAX parser:
下面是演示如何获取JAXB (JSR-222)实现以使用 SAX 解析器的示例代码：
import java.io.FileReader;
import javax.xml.XMLConstants;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.Unmarshaller;
import javax.xml.parsers.SAXParserFactory;
import javax.xml.transform.sax.SAXSource;
import org.xml.sax.InputSource;
import org.xml.sax.XMLReader;
public class Demo {
public static void main(String[] args) throws Exception {
JAXBContext jc = JAXBContext.newInstance(Foo.class);
SAXParserFactory spf = SAXParserFactory.newInstance();
spf.setFeature(XMLConstants.FEATURE_SECURE_PROCESSING, true);
XMLReader xmlReader = spf.newSAXParser().getXMLReader();
InputSource inputSource = new InputSource(new FileReader("input.xml"));
SAXSource source = new SAXSource(xmlReader, inputSource);
Unmarshaller unmarshaller = jc.createUnmarshaller();
Foo foo = (Foo) unmarshaller.unmarshal(source);
System.out.println(foo.getValue());
}
}
回答by Renaud
Building on the answers from @blaise-doughan and @aerobiotic, here is a solution that worked for me:
基于@blaise-doughan 和@aerobiotic 的回答，这里有一个对我有用的解决方案：
import java.io.FileReader;
import javax.xml.XMLConstants;
import javax.xml.bind.JAXBContext;
import javax.xml.bind.Unmarshaller;
import javax.xml.parsers.SAXParserFactory;
import javax.xml.transform.sax.SAXSource;
import org.xml.sax.InputSource;
import org.xml.sax.XMLReader;
public class Demo2 {
public static void main(String[] args) throws Exception {
JAXBContext jc = JAXBContext.newInstance(MyBean.class);
SAXParserFactory spf = SAXParserFactory.newInstance();
spf.setFeature("http://apache.org/xml/features/nonvalidating/load-external-dtd", false);
spf.setFeature("http://xml.org/sax/features/validation", false);
XMLReader xmlReader = spf.newSAXParser().getXMLReader();
InputSource inputSource = new InputSource(
new FileReader("myfile.xml"));
SAXSource source = new SAXSource(xmlReader, inputSource);
Unmarshaller unmarshaller = jc.createUnmarshaller();
MyBean foo = (MyBean) unmarshaller.unmarshal(source);
}
}
回答by Philip Helger
You can create the Unmarshaller directly from a javax.xml.transform.sax.SAXSource.
您可以直接从 javax.xml.transform.sax.SAXSource 创建 Unmarshaller。
See the example on this page: http://docs.oracle.com/cd/E17802_01/webservices/webservices/docs/1.6/api/javax/xml/bind/Unmarshaller.html
请参阅此页面上的示例：http: //docs.oracle.com/cd/E17802_01/webservices/webservices/docs/1.6/api/javax/xml/bind/Unmarshaller.html
Than you "only" need to provide your own URIResolver to that SAXSource
比您“仅”需要向该 SAXSource 提供您自己的 URIResolver
回答by Ihor M.
Check out this resource https://java.net/projects/jaxb/lists/users/archive/2003-11/message/60It worked for me when best answer wasn't working.
查看此资源 https://java.net/projects/jaxb/lists/users/archive/2003-11/message/60当最佳答案不起作用时它对我有用。
