java - Json反序列化为URL(拦截)

Question

我在尝试反序列化 URL

时遇到此错误

Caused by: java.net.MalformedURLException: no protocol: www.boo.com
    at java.net.URL.<init>(URL.java:586) ~[na:1.8.0_45]
    at java.net.URL.<init>(URL.java:483) ~[na:1.8.0_45]
    at java.net.URL.<init>(URL.java:432) ~[na:1.8.0_45]
    at com.fasterxml.jackson.databind.deser.std.FromStringDeserializer$Std._deserialize(FromStringDeserializer.java:212) ~[jackson-databind-2.6.2.jar:2.6.2]
    at com.fasterxml.jackson.databind.deser.std.FromStringDeserializer.deserialize(FromStringDeserializer.java:122) ~[jackson-databind-2.6.2.jar:2.6.2]
    at com.fasterxml.jackson.databind.deser.SettableBeanProperty.deserialize(SettableBeanProperty.java:520) ~[jackson-databind-2.6.2.jar:2.6.2]
    at com.fasterxml.jackson.databind.deser.impl.MethodProperty.deserializeAndSet(MethodProperty.java:95) ~[jackson-databind-2.6.2.jar:2.6.2]
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:337) ~[jackson-databind-2.6.2.jar:2.6.2]
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:131) ~[jackson-databind-2.6.2.jar:2.6.2]
    at com.fasterxml.jackson.databind.deser.std.CollectionDeserializer.deserialize(CollectionDeserializer.java:245) ~[jackson-databind-2.6.2.jar:2.6.2]

波乔:

class foo {
   ...
   URL url
   ...
}

如错误所说，缺少协议(protocol)，如果用户未设置协议(protocol)，如何在反序列化之前插入协议(protocol)？

Answer 1

我结合了之前的两个答案:

public class Foo {
   ...
   @JsonDeserialize(using = UrlDeseralizer.class)
   private URL url;
   ...
}

public class UrlDeseralizer extends JsonDeserializer<URL> {

    private Pattern urlPrefix = Pattern.compile("^(https?://|ftp://).*");

    @Override
    public URL deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        ObjectCodec objectCodec = p.getCodec();
        JsonNode node = objectCodec.readTree(p);
        String stringUrl = node.asText();
        if (!urlPrefix.matcher(stringUrl).matches()) {
            return new URL("http://" + stringUrl);
        } else {
            return new URL(stringUrl);
        }
    }

}