Question

我有一个定义为xsd：double的元素。如果我尝试将一个值285放入元素然后我编组它我得到285.0的输出......那很好。但是，如果我输入一个值285292746，我得到2.85292746E8的输出我是元帅。我喜欢这样的东西，以便双输出不会转换为带小数的科学记数法？基本上我想要285292746或2852292746.0000000

java.lang.Double.toString（）对于为xml生成isues的某些值使用“computerized science notation”。

我知道给定值的双重表示很好。但是值是指数格式的事实，我正在研究的系统是接受我的XML但不知道如何处理指数值并导致我的程序无法正常工作。在WSDL或服务器中更改xsd：double类型对我来说是不可行的。我在客户端工作。

我遇到了Jaxb：绑定xsd：double类型。我仍然很难解决问题，以非扩展格式发送值double值。

package com.logger.client 
import javax.xml.bind.annotation.adapters.XmlAdapter;
import javax.xml.bind.DatatypeConverter;

public class JaxbDoubleSerializer  extends XmlAdapter<String, Double>
{


    public Double unmarshal(String value) {
        return ((double)javax.xml.bind.DatatypeConverter.parseDouble(value));
    }

    public String marshal(Double value) {
        if (value == null) {
            return null;
        }
        return (javax.xml.bind.DatatypeConverter.printDouble((double)(double)value));
    }

}

我需要帮助如何使用DoubleSerializer，以便我可以传递没有指数的值。我尝试在我的类MyLogClient.java中使用xmlAdapter Annotation。我需要知道如何解决这种情况。

package com.logger.client 
import javax.xml.ws.BindingProvider;
import javax.xml.bind.JAXBElement;import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;
import javax.xml.bind.annotation.XmlType;
import javax.xml.bind.annotation.XmlValue;


public class MyLogClient 
{
   //Private member fields

  /** Object factory used to create user-and-role specific objects. */
    private static final ObjectFactory _of = new ObjectFactory();

     @XmlJavaTypeAdapter(JaxbDoubleSerializer.class)
   public JAXBElement<Double> msgFileId;

       @XmlJavaTypeAdapter(JaxbDoubleSerializer.class)
   public Double dNumber;


public final void createEntry;
    (
        final String userName,
        final String time,
        final String logMsgStringId,
        final Params logMsgParamsVal,
        final Integer logMessageFieldID
    )

    throws JAXBException
    {

       JAXBElement<String> username = _of.createParamsParam(userName);
      JAXBElement<String> strTime       = _of.createLogRequestTime(time);


        // Build the LogRequest request.
       final LogRequest _LogRequest = _of.createLogRequest();
       _LogRequest.setUserName(userName);
       _LogRequest.setTime(strTime);

        //Following is the main problem
        int nMsgArgs =  285292746;
       dNumber = Double.parseDouble(Integer.toString(nMsgArgs));  
      //After parsing double Value I get dNumber is now  2.85292746E8

      //MsgFile Id is of Type JAXBElement<Double>
      //CreateLogMessageIdentifier takes Double param
      //So the main problem is here..the value of double field in soap request 
      //is being sent in exponential format. I need to send as I explained above
      //285292746.

       msgFileId   = _of.createLogMessageIdentifier(dNumber); 

      JAXBElement<String> strIdVal = _of.createLogMessageFileId(logMsgStringId);

      final LogMessage logMessage = _of.createLogMessage();

     JAXBElement<Params> _logMsgParams =_of.createLogMessageParams(logMsgParamsVal);

     //Following is where I am trying to use marshall for double type.

     JAXBContext context = JAXBContext.newInstance("package com.logger.client ");
     context.createMarshaller().marshal(msgFileId, System.out);

      logMessage.setIdentifier(msgFileId);  //Method takes JAXBElement<Double>

      logMessage.setFileId(strIdVal );
      logMessage.setParams(_logMsgParams);

      JAXBElement<LogMessage> logMsgValue = _of.createLogRequestLogMessage(logMessage);


        _LogRequest.setLogMessage(logMsgValue);

        // Set the log entry
        port.log(_LogRequest); //Request is sent to server.

}

WSDL xsd：类型声明如下： -

  <xsd:complexType name="LogMessage">
         <xsd:sequence>
        <xsd:element name="fileId" type="xsd:string" minOccurs="0" nillable="true" />
        <xsd:element name="identifier" type="xsd:double" minOccurs="0" nillable="true" />
        <xsd:element name="params" type="tns:Params" minOccurs="0" nillable="true" />
         </xsd:sequence>
           </xsd:complexType>

标识符字段的输出为： -

     <identifier> 2.85292746E8</indentifier> 

Whereas I want to send as. Because system does accept/recognize following types.

<identifier> 285292746</indentifier> 
or 
<identifier> 285292746.00000000</indentifier>

Answer 1

您可以使用如下所示的外部绑定文件：

<强> binding.xml

<jxb:bindings 
    xmlns:xs="http://www.w3.org/2001/XMLSchema"
    xmlns:jxb="http://java.sun.com/xml/ns/jaxb"
    version="2.1">

    <jxb:globalBindings>
            <jxb:javaType name="java.lang.Double"
                xmlType="xs:double"
                parseMethod="javax.xml.bind.DatatypeConverter.parseDouble"
                printMethod="javax.xml.bind.DatatypeConverter.printDouble" />
    </jxb:globalBindings>

</jxb:bindings>

<强> root.xsd

<?xml version="1.0"?>
<xs:schema 
    elementFormDefault="qualified"
    targetNamespace="http://www.example.com"
    xmlns:xs="http://www.w3.org/2001/XMLSchema">

    <xs:complexType name="root">
        <xs:sequence>
            <xs:element name="foo" type="xs:double" />
            <xs:element name="bar" type="xs:double" />
        </xs:sequence>
    </xs:complexType>

</xs:schema>

XJC致电

xjc -d out -b binding.xml root.xsd

<强>根

为根类型生成的类将在Double属性上注册XmlAdapter个类：

package com.example;

import javax.xml.bind.annotation.XmlAccessType;
import javax.xml.bind.annotation.XmlAccessorType;
import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlSchemaType;
import javax.xml.bind.annotation.XmlType;
import javax.xml.bind.annotation.adapters.XmlJavaTypeAdapter;
import org.w3._2001.xmlschema.Adapter1;

@XmlAccessorType(XmlAccessType.FIELD)
@XmlType(name = "root", propOrder = {
    "foo",
    "bar"
})
public class Root {

    @XmlElement(required = true, type = String.class)
    @XmlJavaTypeAdapter(Adapter1 .class)
    @XmlSchemaType(name = "double")
    protected Double foo;
    @XmlElement(required = true, type = String.class)
    @XmlJavaTypeAdapter(Adapter1 .class)
    @XmlSchemaType(name = "double")
    protected Double bar;

    public Double getFoo() {
        return foo;
    }

    public void setFoo(Double value) {
        this.foo = value;
    }

    public Double getBar() {
        return bar;
    }

    public void setBar(Double value) {
        this.bar = value;
    }

}

<强>适配器1

XmlAdapter使用您在binding.xml文件中配置的方法。

package org.w3._2001.xmlschema;

import javax.xml.bind.annotation.adapters.XmlAdapter;

public class Adapter1 extends XmlAdapter<String, Double> {

    public Double unmarshal(String value) {
        return (javax.xml.bind.DatatypeConverter.parseDouble(value));
    }

    public String marshal(Double value) {
        if (value == null) {
            return null;
        }
        return (javax.xml.bind.DatatypeConverter.printDouble(value));
    }

}

Answer 2

最好的方法可能是在JAXB编译器中使用自定义绑定（javaType子句）。可以找到文档here。根据你是否必须编组和解组双打，解决方案可能就像覆盖全局级别的BigDecimal之类的映射一样简单（虽然理解虽然双精度和小数是完全不同的东西，但你似乎想要一个小数表示），或使用数据类型转换器（上面的链接为您提供了所有这些选项的示例）。

Jaxb：使用XMLAdapter进行全局绑定双类型转换

2 个答案: