java - 使用 Jackson 序列化 yml 文件的数据对象

Question

我正在尝试序列化一个 .yml 文件，但我收到消息:

Caused by: com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException: 无法识别的字段“text”(类 com.core.storage.Content)，未标记为可忽略(2 个已知属性:“check”， “标签”])

如消息所示，我不想忽略属性“check”和“label”。这是我的数据对象:

class CheckList(
        index: Int, 
        @JsonProperty("list") content: MutableList<Content> = arrayListOf())

@JsonPropertyOrder("check", "label")
class Content(@JsonProperty("check") val check: String = "",
              @JsonProperty("label") val label: String = "")

序列化函数:

fun <T : Any> parseYmlFile(file: File, c: KClass<T>): T {
    val mapper = ObjectMapper(YAMLFactory())
    mapper.registerModule(KotlinModule())
    return file.bufferedReader().use { mapper.readValue(it.readText(), c.java) }
}

比parseYmlFile(file, CheckList::class)

我在这里错过了什么？我尝试删除 @JsonPropertyOrder 但没有成功

YML:

index: 100
list:
- check: Avoid regular phone calls for sensitive conversations
- check: Use Signal for secure calls
- check: Use JitsiMeet for video conferencing
- label: If you must use less secure options
- check: Download from official website
- check: Change password regularly
- check: Adjust settings so you don't keep chat history
- check: Verify who you're speaking to
- check: Consider using anonymous username
- check: Use codes for sensitive topics

Answer 1

确保对属性使用“val”:

class CheckList(
    val index: Int,
    @JsonProperty("list") val content: MutableList<Content> = arrayListOf())