我有一个对象 - > XML - >我必须支持的一个项目中的对象进程。 该对象包含List,如果它被序列化,则省略列表中存在的所有空值。 我的问题是,可以用Simpleframework完成,还是应该使用其他东西?什么? 这是我的工作:
import java.io.StringWriter;
import java.util.Arrays;
import java.util.List;
import org.simpleframework.xml.Attribute;
import org.simpleframework.xml.ElementList;
import org.simpleframework.xml.Root;
import org.simpleframework.xml.core.Persister;
import org.testng.annotations.Test;
public class SimpleframeworkTest {
@Test
public void testNullsInParams() throws Exception {
Container container = new Container();
container.setId(4000);
container.setParams(Arrays.asList(new Object[] { "foo", null, "bar" }));
String xml = container.toXml(); // omits null value in output
}
@Test
public void testDeserializeNull() throws Exception {
String xml = "<container id=\"4000\">"+
" <object class=\"java.lang.String\">foo</object>"+
// " <object class=\"java.lang.String\"></object>"+ // gets NullPointerException here
" <object class=\"java.lang.String\">bar</object>"+
"</container>";
Container object = Container.toObject(xml);
}
@Root(name = "container", strict = false)
public static class Container {
@Attribute
private Integer id;
@ElementList(inline = true, required = false)
private List<Object> params;
public String toXml() throws Exception {
StringWriter sw = new StringWriter();
new Persister().write(this, sw);
return sw.toString();
}
public static Container toObject(String xml) throws Exception {
return new Persister().read(Container.class, xml);
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public List<Object> getParams() {
return params;
}
public void setParams(List<Object> params) {
this.params = params;
}
@Override
public String toString() {
return "Container [id=" + id + ", params=" + params + "]";
}
}
}
答案 0 :(得分:1)
首先,列表注释缺少条目名称:
@ElementList(inline = true, required = false, entry = "object")
private List<Object> params;
否则使用<string>...</string>
,而不是<object>...</object>
您可以通过将type = String.class
添加到列表的注释来阻止该nullpointer excpetion。但是,这并不能解决主要问题。
通常空标记/ null
- 元素不会添加到结果中。
以下是如何使用Converter
来解决此问题的示例。
public class SimpleframeworkTest
{
// ...
@Root(name = "container", strict = false)
@Convert(NullawareContainerConverter.class)
public static class Container
{
static final Serializer ser = new Persister(new AnnotationStrategy());
// ...
public String toXml() throws Exception
{
StringWriter sw = new StringWriter();
ser.write(this, sw);
return sw.toString();
}
public static Container toObject(String xml) throws Exception
{
return ser.read(Container.class, xml);
}
// ...
}
static class NullawareContainerConverter implements Converter<Container>
{
final Serializer ser = new Persister();
@Override
public Container read(InputNode node) throws Exception
{
final Container c = new Container();
c.id = Integer.valueOf(node.getAttribute("id").getValue());
c.params = new ArrayList<>();
InputNode n;
while( ( n = node.getNext("object")) != null )
{
/*
* If the value is null it's added too. You also can add some
* kind of null-replacement element here too.
*/
c.params.add(n.getValue());
}
return c;
}
@Override
public void write(OutputNode node, Container value) throws Exception
{
ser.write(value.id, node);
for( Object obj : value.params )
{
if( obj == null )
{
obj = ""; // Set a valid value if null
}
// Possible you have to tweak this by hand
ser.write(obj, node);
}
}
}
}
正如评论中所写,你必须做进一步的工作。
<强>结果:强>
testNullsInParams()
<container>
<integer>4000</integer>
<string>foo</string>
<string></string>
<string>bar</string>
</container>
testDeserializeNull()
Container [id=4000, params=[foo, null, bar]]