Question

我有一些对象是由JAXB从XML文件解组的。是否有可能让JAXB告诉我或以某种方式找出每个对象来自XML文件（行和列）的位置？

此信息在某些时候可用，因为JAXB在架构验证错误期间将其提供给我。但我也希望它可用于经过验证的对象。

Answer 1

您可以通过利用XMLStreamReader和Unmarshaller.Listener在JAXB中执行此操作：

<强>演示

package forum383861;

import java.io.FileInputStream;
import java.util.HashMap;
import java.util.Map;

import javax.xml.bind.JAXBContext;
import javax.xml.bind.Unmarshaller;
import javax.xml.bind.Unmarshaller.Listener;
import javax.xml.stream.Location;
import javax.xml.stream.XMLInputFactory;
import javax.xml.stream.XMLStreamReader;

public class Demo {

    public static void main(String[] args) throws Exception {
        JAXBContext jc = JAXBContext.newInstance(Customer.class);


        XMLInputFactory xif = XMLInputFactory.newFactory();
        FileInputStream xml = new FileInputStream("src/forum383861/input.xml");
        XMLStreamReader xsr = xif.createXMLStreamReader(xml);

        Unmarshaller unmarshaller = jc.createUnmarshaller();
        LocationListener ll = new LocationListener(xsr);
        unmarshaller.setListener(ll);

        Customer customer = (Customer) unmarshaller.unmarshal(xsr);
        System.out.println(ll.getLocation(customer));
        System.out.println(ll.getLocation(customer.getAddress()));
    }

    private static class LocationListener extends Listener {

        private XMLStreamReader xsr;
        private Map<Object, Location> locations;

        public LocationListener(XMLStreamReader xsr) {
            this.xsr = xsr;
            this.locations = new HashMap<Object, Location>();
        }

        @Override
        public void beforeUnmarshal(Object target, Object parent) {
            locations.put(target, xsr.getLocation());
        }

        public Location getLocation(Object o) {
            return locations.get(o);
        }

    }

}

<强> input.xml中

<?xml version="1.0" encoding="UTF-8"?>
<customer>
    <address/>
</customer>

<强>输出

[row,col {unknown-source}]: [2,1]
[row,col {unknown-source}]: [3,5]

<强>客户

package forum383861;

import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement
public class Customer {

    private Address address;

    public Address getAddress() {
        return address;
    }

    public void setAddress(Address address) {
        this.address = address;
    }

}

<强>地址

package forum383861;

public class Address {

}

了解更多信息

http://blog.bdoughan.com/2011/08/using-unmarshallerlistener-to-capture.html

Answer 2

我不敢。 JAXB构建在XML解析器之上，这个解析器将构建XML文档的逻辑表示，忘记文档的原始字符串表示。

验证步骤在您的字符串仍然被读入时完成，因此您的解析器能够向您提供一条错误消息，告诉您错误的位置。 JAXB只会绕过该错误消息。但是只要验证和解析XML，就只存在逻辑表示。

用于解组对象的文件中的JAXB位置

2 个答案: