How to deserialize a float value with a localized

Question

The input stream I am parsing with Jackson contains latitude and longitude values such as here:

{
    "name": "product 23",
    "latitude": "52,48264",
    "longitude": "13,31822"
}

For some reason the server uses commas as the decimal separator which produces an InvalidFormatException. Since I cannot change the server output format I would like to teach Jackson's ObjectMapper to handle those cases. Here is the relevant code:

public static Object getProducts(final String inputStream) {
    ObjectMapper objectMapper = new ObjectMapper();
    try {
        return objectMapper.readValue(inputStream,
                new TypeReference<Product>() {}
        );
    } catch (UnrecognizedPropertyException e) {
        e.printStackTrace();
    } catch (InvalidFormatException e) {
        e.printStackTrace();
    } catch (JsonMappingException e) {
        e.printStackTrace();
    } catch (JsonParseException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    return null;
}

And here is the POJO:

import com.fasterxml.jackson.annotation.JsonProperty;

public class Product {

    @JsonProperty("name")
    public String name;
    @JsonProperty("latitude")
    public float latitude;
    @JsonProperty("longitude")
    public float longitude;

}

How can I tell Jackson that those coordinate values come with a German locale?

---

I suppose a custom deserializer for the specific fields as discussed here would be the way to go. I drafted this:

public class GermanFloatDeserializer extends JsonDeserializer<Float> {

    @Override
    public Float deserialize(JsonParser parser, DeserializationContext context)
            throws IOException {
        // TODO Do some comma magic
        return floatValue;
    }

}

Then the POJO would look like this:

import com.fasterxml.jackson.annotation.JsonProperty;

public class Product {

    @JsonProperty("name")
    public String name;
    @JsonDeserialize(using = GermanFloatDeserializer.class, as = Float.class)
    @JsonProperty("latitude")
    public float latitude;
    @JsonDeserialize(using = GermanFloatDeserializer.class, as = Float.class)
    @JsonProperty("longitude")
    public float longitude;

}

Answer 1

With all respect to accepted answer, there is a way to get rid of those @JsonDeserialize annotations.

You need to register the custom deserializer in the ObjectMapper.

Following the tutorial from official web-site you just do something like:

    ObjectMapper mapper = new ObjectMapper();
    SimpleModule testModule = new SimpleModule(
            "DoubleCustomDeserializer",
            new com.fasterxml.jackson.core.Version(1, 0, 0, null))
            .addDeserializer(Double.class, new JsonDeserializer<Double>() {
                @Override
                public Double deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException {
                    String valueAsString = jp.getValueAsString();
                    if (StringUtils.isEmpty(valueAsString)) {
                        return null;
                    }

                    return Double.parseDouble(valueAsString.replaceAll(",", "\\."));
                }
            });
    mapper.registerModule(testModule);

If you're using Spring Boot there is a simpler method. Just define the Jackson2ObjectMapperBuilder bean somewhere in your Configuration class:

@Bean
public Jackson2ObjectMapperBuilder jacksonBuilder() {
    Jackson2ObjectMapperBuilder builder = new Jackson2ObjectMapperBuilder();

    builder.deserializerByType(Double.class, new JsonDeserializer<Double>() {
        @Override
        public Double deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException {
            String valueAsString = jp.getValueAsString();
            if (StringUtils.isEmpty(valueAsString)) {
                return null;
            }

            return Double.parseDouble(valueAsString.replaceAll(",", "\\."));
        }
    });

    builder.applicationContext(applicationContext);
    return builder;
}

and add the custom HttpMessageConverter to the list of WebMvcConfigurerAdapter message converters:

 messageConverters.add(new MappingJackson2HttpMessageConverter(jacksonBuilder().build()));

Answer 2

A more general solution than the other proposed answers, which require registering individual deserializers for each type, is to provide a customized DefaultDeserializationContext to ObjectMapper.

The following implementation (which is inspired by DefaultDeserializationContext.Impl) worked for me:

class LocalizedDeserializationContext extends DefaultDeserializationContext {
    private final NumberFormat format;

    public LocalizedDeserializationContext(Locale locale) {
        // Passing `BeanDeserializerFactory.instance` because this is what happens at
        // 'jackson-databind-2.8.1-sources.jar!/com/fasterxml/jackson/databind/ObjectMapper.java:562'.
        this(BeanDeserializerFactory.instance, DecimalFormat.getNumberInstance(locale));
    }

    private LocalizedDeserializationContext(DeserializerFactory factory, NumberFormat format) {
        super(factory, null);
        this.format = format;
    }

    private LocalizedDeserializationContext(DefaultDeserializationContext src, DeserializationConfig config, JsonParser parser, InjectableValues values, NumberFormat format) {
        super(src, config, parser, values);
        this.format = format;
    }

    @Override
    public DefaultDeserializationContext with(DeserializerFactory factory) {
        return new LocalizedDeserializationContext(factory, format);
    }

    @Override
    public DefaultDeserializationContext createInstance(DeserializationConfig config, JsonParser parser, InjectableValues values) {
        return new LocalizedDeserializationContext(this, config, parser, values, format);
    }

    @Override
    public Object handleWeirdStringValue(Class<?> targetClass, String value, String msg, Object... msgArgs) throws IOException {
        // This method is called when default deserialization fails.
        if (targetClass == float.class || targetClass == Float.class) {
            return parseNumber(value).floatValue();
        }
        if (targetClass == double.class || targetClass == Double.class) {
            return parseNumber(value).doubleValue();
        }
        // TODO Handle `targetClass == BigDecimal.class`?
        return super.handleWeirdStringValue(targetClass, value, msg, msgArgs);
    }

    // Is synchronized because `NumberFormat` isn't thread-safe.
    private synchronized Number parseNumber(String value) throws IOException {
        try {
            return format.parse(value);
        } catch (ParseException e) {
            throw new IOException(e);
        }
    }
}

Now set up your object mapper with your desired locale:

Locale locale = Locale.forLanguageTag("da-DK");
ObjectMapper objectMapper = new ObjectMapper(null,
                                             null,
                                             new LocalizedDeserializationContext(locale));

If you use Spring RestTemplate, you can set it up to use objectMapper like so:

RestTemplate template = new RestTemplate();
template.setMessageConverters(
    Collections.singletonList(new MappingJackson2HttpMessageConverter(objectMapper))
);

Note that the value must be represented as a string in the JSON document (i.e. {"number": "2,2"}), since e.g. {"number": 2,2} is not valid JSON and will fail to parse.

Answer 3

I came up with the following solution:

public class FlexibleFloatDeserializer extends JsonDeserializer<Float> {

    @Override
    public Float deserialize(JsonParser parser, DeserializationContext context)
            throws IOException {
        String floatString = parser.getText();
        if (floatString.contains(",")) {
            floatString = floatString.replace(",", ".");
        }
        return Float.valueOf(floatString);
    }

}

...

public class Product {

    @JsonProperty("name")
    public String name;
    @JsonDeserialize(using = FlexibleFloatDeserializer.class)
    @JsonProperty("latitude")
    public float latitude;
    @JsonDeserialize(using = FlexibleFloatDeserializer.class)
    @JsonProperty("longitude")
    public float longitude;

}

Still I wonder why I it does not work when I specify the return value class as as = Float.class as can be found in the documentation of JsonDeserialize. It reads as if I am supposed to use one or the other but not both. Whatsoever, the docs also claim that as = will be ignored when using = is defined:

if using() is also used it has precedence (since it directly specified deserializer, whereas this would only be used to locate the deserializer) and value of this annotation property is ignored.