如何在Kotlin中从给定JSON字符串获取属性名列表

方法一：使用Java标准库的org.json（无需额外依赖） #

Android项目默认包含该库，JVM项目可手动引入依赖，实现代码如下：

import org.json.JSONObject

fun main() {
    val jsonString = """
        {
            "id": 2,
            "title": "iPhone X",
            "description": "SIM-Free, Model A19211 6.5-inch Super Retina HD display with OLED technology A12 Bionic chip with ...",
            "price": 899,
            "discountPercentage": 17.94,
            "rating": 4.44,
            "stock": 34,
            "brand": "Apple",
            "category": "smartphones",
            "thumbnail": "https://i.dummyjson.com/data/products/2/thumbnail.jpg",
            "images": [
                "https://i.dummyjson.com/data/products/2/1.jpg",
                "https://i.dummyjson.com/data/products/2/2.jpg",
                "https://i.dummyjson.com/data/products/2/3.jpg",
                "https://i.dummyjson.com/data/products/2/thumbnail.jpg"
            ]
        }
    """.trimIndent()

    val jsonObject = JSONObject(jsonString)
    val propertyNames = jsonObject.keys().asSequence().toList()
    println(propertyNames)
}

方法二：使用Gson库 #

若项目用Gson处理JSON，可通过JsonObject直接获取属性名：

import com.google.gson.JsonParser

fun main() {
    val jsonString = """
        {
            "id": 2,
            "title": "iPhone X",
            "description": "SIM-Free, Model A19211 6.5-inch Super Retina HD display with OLED technology A12 Bionic chip with ...",
            "price": 899,
            "discountPercentage": 17.94,
            "rating": 4.44,
            "stock": 34,
            "brand": "Apple",
            "category": "smartphones",
            "thumbnail": "https://i.dummyjson.com/data/products/2/thumbnail.jpg",
            "images": [
                "https://i.dummyjson.com/data/products/2/1.jpg",
                "https://i.dummyjson.com/data/products/2/2.jpg",
                "https://i.dummyjson.com/data/products/2/3.jpg",
                "https://i.dummyjson.com/data/products/2/thumbnail.jpg"
            ]
        }
    """.trimIndent()

    val jsonObject = JsonParser.parseString(jsonString).asJsonObject
    val propertyNames = jsonObject.keySet().toList()
    println(propertyNames)
}

方法三：使用Jackson库 #

对于依赖Jackson的项目，通过JsonNode提取属性名：

import com.fasterxml.jackson.databind.JsonNode
import com.fasterxml.jackson.databind.ObjectMapper

fun main() {
    val jsonString = """
        {
            "id": 2,
            "title": "iPhone X",
            "description": "SIM-Free, Model A19211 6.5-inch Super Retina HD display with OLED technology A12 Bionic chip with ...",
            "price": 899,
            "discountPercentage": 17.94,
            "rating": 4.44,
            "stock": 34,
            "brand": "Apple",
            "category": "smartphones",
            "thumbnail": "https://i.dummyjson.com/data/products/2/thumbnail.jpg",
            "images": [
                "https://i.dummyjson.com/data/products/2/1.jpg",
                "https://i.dummyjson.com/data/products/2/2.jpg",
                "https://i.dummyjson.com/data/products/2/3.jpg",
                "https://i.dummyjson.com/data/products/2/thumbnail.jpg"
            ]
        }
    """.trimIndent()

    val objectMapper = ObjectMapper()
    val jsonNode: JsonNode = objectMapper.readTree(jsonString)
    val propertyNames = jsonNode.fieldNames().asSequence().toList()
    println(propertyNames)
}

以上三种方法最终都会输出期望的结果：

[id, title, description, price, discountPercentage, rating, stock, brand, category, thumbnail, images]

内容的提问来源于stack exchange，提问作者Al Masud