Android - 将 XML 解析为 JSON

Question

我的用例要求我获取格式如下的 XML 并需要将其映射到 JSON 对象：

<settings>
<client ip="......" lat="...." lon="...." isp="...." country="...."/>
<translation lang="xml"> </translation>
</settings>

我正在使用 Retrofit 和 Moshi 作为请求的转换器工厂：

       Retrofit.Builder()
            .addConverterFactory(MoshiConverterFactory.create(moshi).asLenient())
            .baseUrl(BASE_URL)
            .client(okHttpClient)
            .build()

我尝试在 SimpleXml (com.squareup.retrofit2:converter-simplexml:2.9.0) 的帮助下创建响应模型，但到目前为止还没有成功。我不断收到此错误：

com.squareup.moshi.JsonDataException：应为 BEGIN_OBJECT，但在路径 $ 处为 STRING

任何人都可能知道如何将此结构映射到 Json？提前谢谢！

响应模型：

@Root(name = "settings", strict = false)
data class ConfigurationEntity(

    @field:Attribute(name = "client", required = false)
    @param:Attribute(name = "client", required = false)
    val client: ClientEntity?
)

@Root(name = "client")
data class ClientEntity(

    @field:Attribute(name = "ip", required = false)
    @param:Attribute(name = "ip", required = false)
    val ip: String?,

    @field:Attribute(name = "lat", required = false)
    @param:Attribute(name = "lat", required = false)
    val lat: String?,

    @field:Attribute(name = "lon", required = false)
    @param:Attribute(name = "lon", required = false)
    val lon: String?,

    @field:Attribute(name = "isp", required = false)
    @param:Attribute(name = "isp", required = false)
    val isp: String?,

    @field:Attribute(name = "country", required = false)
    @param:Attribute(name = "country", required = false)
    val country: String?
)

解决方案

改造建造者：

       Retrofit.Builder()
            .addConverterFactory(SimpleXmlConverterFactory.create())
            .baseUrl(BASE_URL)
            .client(okHttpClient)
            .build()

型号：

@Root(name = "settings", strict = false)
class ConfigurationEntity {

    @field:Element(required = false)
    lateinit var client: ClientEntity
}

@Root(name = "client", strict = false)
class ClientEntity {

    @field:Attribute
    lateinit var ip: String

    @field:Attribute
    lateinit var lat: String

    @field:Attribute
    lateinit var lon: String

    @field:Attribute
    lateinit var isp: String

    @field:Attribute
    lateinit var country: String
}

这个链接也可能有帮助Kotlin - parsing Xml response with list using SimpleXml

Answer 1

您需要将此转换器工厂添加到您的 Retrofit Builder -

Retrofit.Builder()
.addConverterFactory(SimpleXmlConverterFactory.create())

这允许 Retrofit 解释它正在获取的 XML 以转换为它的等效 JSON 数据对象。