java – Jackson根据json输入映射Object或Object列表
作者:互联网
我有这个POJO:
public class JsonObj {
private String id;
private List<Location> location;
public String getId() {
return id;
}
public List<Location> getLocation() {
return location;
}
@JsonSetter("location")
public void setLocation(){
List<Location> list = new ArrayList<Location>();
if(location instanceof Location){
list.add((Location) location);
location = list;
}
}
}
json输入中的“location”对象可以是Location的简单实例,也可以是Location的Array.当它只是一个实例时,我收到此错误:
Could not read JSON: Can not deserialize instance of java.util.ArrayList out of START_OBJECT token
我试图实现自定义setter但它不起作用.我怎么能根据json输入映射位置或列表?
解决方法:
更新:Mher Sarkissian’s soulution工作正常,它也可以使用建议here注释,如下所示:
@JsonFormat(with = JsonFormat.Feature.ACCEPT_SINGLE_VALUE_AS_ARRAY)
private List<Item> item;
我对这个最烦人的问题表示最深切的同情,我遇到了同样的问题并在此找到了解决方案:https://stackoverflow.com/a/22956168/1020871
通过一点修改,我想出了这个,首先是泛型类:
public abstract class OptionalArrayDeserializer<T> extends JsonDeserializer<List<T>> {
private final Class<T> clazz;
public OptionalArrayDeserializer(Class<T> clazz) {
this.clazz = clazz;
}
@Override
public List<T> deserialize(JsonParser jp, DeserializationContext ctxt)
throws IOException {
ObjectCodec oc = jp.getCodec();
JsonNode node = oc.readTree(jp);
ArrayList<T> list = new ArrayList<>();
if (node.isArray()) {
for (JsonNode elementNode : node) {
list.add(oc.treeToValue(elementNode, clazz));
}
} else {
list.add(oc.treeToValue(node, clazz));
}
return list;
}
}
然后属性和实际的反序列化器类(Java泛型并不总是很漂亮):
@JsonDeserialize(using = ItemListDeserializer.class)
private List<Item> item;
public static class ItemListDeserializer extends OptionalArrayDeserializer<Item> {
protected ItemListDeserializer() {
super(Item.class);
}
}
标签:android,java,json,jackson,rest-client 来源: https://codeday.me/bug/20190930/1836247.html