小编Moh*_*ham的帖子

我正在尝试解组 xml 文件。下面是我的文件。

XML文件

<AccountInfo>
   <Account>
    <FirstName>FirstName1</FirstName>
    <LastName>LastName1</LastName>
    <Country>US</Country>
    <PhoneNumber>2022023111</PhoneNumber>
    <Email>test@test.com</Email>
    <Password>1222433123</Password>
    <PaymentMethod>
        <CardNumber>4111111111111111</CardNumber>
        <Month>12</Month>
        <Year>2017</Year>
        <CVV>098</CVV>
        <StreetAddress>test</StreetAddress>
        <Town>test</Town>
        <State>Virginia</State>
        <ZipCode>12365</ZipCode>
        <Country>US</Country>
    </PaymentMethod>
 </Account>
 <Account>
    <FirstName>FirstName1</FirstName>
    <LastName>LastName1</LastName>
    <Country>US</Country>
    <PhoneNumber>2022023111</PhoneNumber>
    <Email>test@test.com</Email>
    <Password>1222433123</Password>
    <PaymentMethod>
        <CardNumber>4111111111111111</CardNumber>
        <Month>12</Month>
        <Year>2017</Year>
        <CVV>098</CVV>
        <StreetAddress>test</StreetAddress>
        <Town>test</Town>
        <State>Virginia</State>
        <ZipCode>12365</ZipCode>
        <Country>US</Country>
    </PaymentMethod>
 </Account>
</AccountInfo>

JaxbHelper.java

 public static <T> T unmarshal(InputStream file, Class<T> unmarshalClass) 
 throws JAXBException {
    JAXBContext jaxbContext = JAXBContext.newInstance(unmarshalClass);
    Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
    T object = (T) jaxbUnmarshaller.unmarshal(file);
    return object;
}

账户信息.java

  @XmlRootElement(name = "AccountInfo")
  @XmlAccessorType(XmlAccessType.FIELD)
  @XmlSeeAlso({PaymentMethod.class})
  public class …

通过MVC 5 HTML上传文件时是否可以完全删除"所有文件"下拉选项

@TempData ["fileType"]接受(.xml,.txt,.xml)它可以工作,但我想完全删除所有文件

 <input type="file" name="file" accept="@TempData["fileType"]"  />

提前致谢.