Question

JsonGenerator generator = 
                new JsonFactory().createJsonGenerator(new JSONWriter(response));
generator.configure(JsonGenerator.Feature.WRITE_NUMBERS_AS_STRINGS, true);

我使用JsonGenerator.Feature.WRITE_NUMBERS_AS_STRINGS在json中将数字写为字符串。但是，我找不到类似的功能来将布尔值写为字符串。

Answer 1

我也找不到类似的布尔特征。因此，我建议为布尔字段编写新的序列化器和反序列化器。

参见我的例子：

public class JacksonProgram {

    public static void main(String[] args) throws IOException {
        Foo foo = new Foo();
        foo.setB(true);
        foo.setS("Test");
        foo.setI(39);

        ObjectMapper objectMapper = new ObjectMapper();
        JsonFactory jsonFactory = new JsonFactory();

        StringWriter stringWriter = new StringWriter();
        JsonGenerator jsonGenerator = jsonFactory.createGenerator(stringWriter);
        jsonGenerator.enable(JsonGenerator.Feature.WRITE_NUMBERS_AS_STRINGS);
        objectMapper.writeValue(jsonGenerator, foo);
        System.out.println(stringWriter);

        JsonParser jsonParser = jsonFactory.createJsonParser(stringWriter.toString());
        Foo value = objectMapper.readValue(jsonParser, Foo.class);
        System.out.println(value);
    }
}

class BooleanSerializer extends JsonSerializer<Boolean> {

    @Override
    public void serialize(Boolean value, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws IOException, JsonProcessingException {
        jsonGenerator.writeString(value.toString());
    }
}

class BooleanDeserializer extends JsonDeserializer<Boolean> {

    public Boolean deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException, JsonProcessingException {
        return Boolean.valueOf(jsonParser.getValueAsString());
    }
}

class Foo {

    @JsonSerialize(using = BooleanSerializer.class)
    @JsonDeserialize(using = BooleanDeserializer.class)
    private boolean b;
    private String s;
    private int i;

    public boolean isB() {
        return b;
    }

    public void setB(boolean b) {
        this.b = b;
    }

    public String getS() {
        return s;
    }

    public void setS(String s) {
        this.s = s;
    }

    public int getI() {
        return i;
    }

    public void setI(int i) {
        this.i = i;
    }

    @Override
    public String toString() {
        return "Foo [b=" + b + ", s=" + s + ", i=" + i + "]";
    }
}

输出：

{"b":"true","s":"Test","i":"39"}
Foo [b=true, s=Test, i=39]

修改

我认为，您应该将SimpleModule配置添加到ObjectMapper：

SimpleModule simpleModule = new SimpleModule("BooleanModule"); simpleModule.addSerializer(Boolean.class, new BooleanSerializer()); simpleModule.addDeserializer(Boolean.class, new BooleanDeserializer()); ObjectMapper objectMapper = new ObjectMapper(); objectMapper.registerModule(simpleModule);

现在，你应该能够序列化boolean / Object List-s和Map-s。

Answer 2

我所知道的工作json字符串看起来像这样：

rcffull$RetBackers <- as.numeric(as.character(rcffull$Returning.Backers))

rcffull$NewBackers <- as.numeric(as.character(rcffull$New.Backers))

使用特定于您使用的语言的多行语法。

如何在json数组中将布尔值写为String？

2 个答案: