实用小工具： Kotlin 代码片段

1.将任意对象转换成Map

inline fun <reified T : Any> T.asMap() : Map<String, Any?> {
    
    val props = T::class.memberProperties.associateBy {
     it.name }
    return props.keys.associateWith {
     props[it]?.get(this) }
}

2. 实用ObjectMapper转换对象

2.1 对象转JsonString

data class Person(val name: String, var age: Int)

fun obj2JsonString(obj: Person){
    
	val objectMapper =   ObjectMapper().registerModule(JavaTimeModule()).disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES)
            .disable(SerializationFeature.FAIL_ON_EMPTY_BEANS)
return objectMapper.writeValueAsString(person)
}

2.2 JsonString转对象

data class Person(val name: String = " ", var age: Int = 23)

fun getPersonFromJsonString(): Person {
    
	val objectMapper =   ObjectMapper().registerModule(JavaTimeModule()).disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES)
            .disable(SerializationFeature.FAIL_ON_EMPTY_BEANS)
val jsonString = "{\"name\":\"tom\",\"age\":34}"
return objectMapper.readValue(jsonString, Person::class.java)
}

注意：
可以看到，这里的data class的每个字段都给了默认值得，如果不给，会报错：

Exception in thread "main" com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `com.example.Person` (no Creators, like default constructor, exist): cannot deserialize from Object value (no delegate- or property-based Creator)
 at [Source: (String)"{"name":"tom","age":34}"; line: 1, column: 2]
	at com.fasterxml.jackson.databind.exc.InvalidDefinitionException.from(InvalidDefinitionException.java:67)
	at com.fasterxml.jackson.databind.DeserializationContext.reportBadDefinition(DeserializationContext.java:1615)
	at com.fasterxml.jackson.databind.DatabindContext.reportBadDefinition(DatabindContext.java:400)
	at com.fasterxml.jackson.databind.DeserializationContext.handleMissingInstantiator(DeserializationContext.java:1077)
	at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.deserializeFromObjectUsingNonDefault(BeanDeserializerBase.java:1332)
	at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:331)
	at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:164)
	at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:4524)
	at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3466)
	at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3434)

这是因为缺少一个无参的构造函数，而data class没有生成无参构造函数，需要为每个参数添加默认值，以产生无参构造函数。kotlin官网上有

给每个data class的每个属性都添加默认值，这多烦人的一件事！你想每个都写吗？反正我不想。
好在，kotlin官方还是有解决的：
使用no-arg的编译插件，在pom文件里作如下配置：

<plugin>
    <artifactId>kotlin-maven-plugin</artifactId>
    <groupId>org.jetbrains.kotlin</groupId>
    <version>${kotlin.version}</version>

    <configuration>
        <compilerPlugins>
            <!-- Or "jpa" for JPA support -->
            <plugin>no-arg</plugin>
        </compilerPlugins>

        <pluginOptions>
            <option>no-arg:annotation=com.my.Annotation</option>
            <!-- Call instance initializers in the synthetic constructor -->
            <!-- <option>no-arg:invokeInitializers=true</option> -->
        </pluginOptions>
    </configuration>

    <dependencies>
        <dependency>
            <groupId>org.jetbrains.kotlin</groupId>
            <artifactId>kotlin-maven-noarg</artifactId>
            <version>${kotlin.version}</version>
        </dependency>
    </dependencies>
</plugin>

2.3 JsonString中包含list

如果json字符串中包含了list，那么就应该写成：

data class Person(val name: String = " ", var age: Int = 23)

fun getPersonFromJsonString(): List<*> {
    
	val objectMapper =   ObjectMapper().registerModule(JavaTimeModule()).disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES)
            .disable(SerializationFeature.FAIL_ON_EMPTY_BEANS)
val jsonString = "[{\"name\":\"tom\",\"age\":34},{\"name\":\"tom\",\"age\":34}]"
    return objectMapper.readValue(jsonString, List::class.java)}

但是这样得到的list中，list中的元素并没有识别出来，而是以linkedHashMap的形式存储在list中的。如果我想取出具体的对象类型，应该这样：

data class Person(val name: String = " ", var age: Int = 23)

fun getPersonFromJsonString(): List<*> {
    
	val objectMapper =   ObjectMapper().registerModule(JavaTimeModule()).disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES)
            .disable(SerializationFeature.FAIL_ON_EMPTY_BEANS)
val jsonString = "[{\"name\":\"tom\",\"age\":34},{\"name\":\"tom\",\"age\":34}]"
    var people: List<Person> = objectMapper.readValue(jsonString, objectMapper.typeFactory.constructCollectionType(List::class.java, Person::class.java))
    return people

这样就可以得到list中包含的具体实例了。