Utility gadget: kotlin code snippet

1. Convert any object to 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. practical ObjectMapper Convert objects

2.1 Object turn 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 Transfer object

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)
}

Be careful ：
You can see , there data class Each field of is given a default value , If not , Will report a mistake ：

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)

This is due to the lack of a parameterless constructor , and data class No parameterless constructor was generated , You need to add default values for each parameter , To generate a parameterless constructor .kotlin Official website There are

For each data class Add default values for each attribute of , What an annoying thing ！ Do you want everyone to write ？ I don't want to .
Fortunately ,kotlin There are still official solutions ：
Use no-arg Compiler plug-in , stay pom The following configuration is made in the file ：

<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 Contained in the list

If json The string contains list, Then it should be written as ：

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)}

But the result is list in ,list Elements in are not recognized , But rather linkedHashMap The form is stored in list Medium . If I want to extract specific object types , It should be ：

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

So you can get list Specific examples included in .