我正在寻找最佳解决方案如何将POJO或
JSON转换为XML,并将所有atttributes转换为正确的位置.目前,杰克逊看起来是最方便的方式.我能够将POJO序列化为没有属性的XML.
POJO TestUser
public class TestUser extends JsonType
{
@JsonProperty("username")
private final String username;
@JsonProperty("fullname")
private final String fullname;
@JsonProperty("email")
private final String email;
@JsonProperty("enabled")
private final Boolean enabled;
@JsonCreator
public TestUser(
@JsonProperty("username") String username,@JsonProperty("fullname") String fullname,@JsonProperty("email") String email,@JsonProperty("enabled") Boolean enabled)
{
this.username = username;
this.fullname = fullname;
this.email = email;
this.enabled = enabled;
}
@JsonGetter("username")
public String getUsername()
{
return username;
}
@JsonGetter("fullname")
public String getFullname()
{
return fullname;
}
@JsonGetter("email")
public String getEmail()
{
return email;
}
@JsonGetter("enabled")
public Boolean getEnabled()
{
return enabled;
}
}
}
这是代码：
public void testJsonToXML() throws JsonParseException,JsonMappingException,IOException
{
String jsonInput = "{\"username\":\"FOO\",\"fullname\":\"FOO BAR\",\"email\":\"foobar@foobar.com\",\"enabled\":\"true\"}";
ObjectMapper jsonMapper = new ObjectMapper();
TestUser foo = jsonMapper.readValue(jsonInput,TestUser.class);
XmlMapper xmlMapper = new XmlMapper();
System.out.println(xmlMapper.writer().with(SerializationFeature.WRAP_ROOT_VALUE).withRootName("product").writeValueAsString(foo));
}
现在它又回来了
<TestUser xmlns="">
<product>
<username>FOO</username>
<fullname>FOO BAR</fullname>
<email>foobar@foobar.com</email>
<enabled>true</enabled>
</product>
</TestUser>
哪个好,但是我需要启用变量作为username的属性然后我需要将xmlns和xsi属性添加到根元素,因此XML结果看起来像这样
<TestUser xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation="testUser.xsd">
<product>
<username enabled="true">FOO</username>
<fullname>FOO BAR</fullname>
<email>foobar@foobar.com</email>
</product>
</TestUser>
我找到了一些使用@JacksonXmlProperty的例子,但它只将属性添加到根元素.
感谢帮助
