Kotlin如何将每一行保存成Class个对象？

Question

我收到来自 API 的列表。我把它按行划分了。我不确定如何将每一行保存到 class 个对象中？你可以帮帮我吗？ [在此处输入图片描述][1]

class RecordsList {


    @RequestMapping("/")
    fun receiveAll(): List<String>? {
        val restTemplate = RestTemplate()
        val url = "some URL // doesn't matter"
        val response = restTemplate.getForObject(url, String::class.java)

        var lines = response?.lines()
        lines?.forEach { line -> println(line)}
        return lines
    }
}

data class Record(var domain: String, var code: String, var link: String, var other: String)

Answer 1

您应该创建一个变量来存储记录的数组列表。

class RecordsList {


@RequestMapping("/")
fun receiveAll(): ArrayList<Record>? {

    // ArrayList to store the `Record`s
    var arrayListOfRecord: ArrayList<Record>? = arrayListOf()

    val restTemplate = RestTemplate()
    val url = "some URL // doesn't matter"
    val response = restTemplate.getForObject(url, String::class.java)

    var lines = response?.lines()

    // For each line in the response create a Record object and add it to the 
    // `arrayListOfRecord`s
    lines?.forEach { line -> 
        arrayListOfRecord?.add(
            Record(
                domain = line.domain,
                code = line.code,
                link = line.link,
                other = line.other
            )
        )
    }
    return arrayListOfRecord 
    }
}

data class Record(var domain: String, var code: String, var link: String, var other: String)

Kotlin如何将每一行保存成Class个对象？

Kotlin how to save each line into Class Objects?

class

object

resttemplate

kotlin

spring-boot