kotlin:2获取404 url未找到错误

正在获取响应{protocol=http/1.1，code=404，message=notfound，url=}

.addConverterFactory(ScalarsConverterFactory.create())

url是正确的，因为邮递员工作正常

.addConverterFactory(ScalarsConverterFactory.create())

我已经试着研究这个错误，但大多数情况下，URL是正确的。错误本身是模糊的

.addConverterFactory(ScalarsConverterFactory.create())

启动它的代码。生成器是一个有效的json字符串。我已经在邮递员身上测试过了

    CoroutineScope(Dispatchers.Default).launch {

        val call = submitService.submitCarton(builder.toString())
        Log.d("submit", "begining")
        withContext(Dispatchers.Main) {
            if (call.isSuccessful) { 
                Log.d("submit",call.body() as String)

            } else {
                Log.d("submit", "else....")

            }
        }
    }

.addConverterFactory(ScalarsConverterFactory.create())

服务工厂：

    fun makeSubmitService() : SubmitService{
    val url =  "https://test.test.com/service/"
    return Retrofit.Builder().baseUrl(url)
        .client(okHttpClient).addConverterFactory(GsonConverterFactory.create())
        .build().create(SubmitService::class.java)
}

.addConverterFactory(ScalarsConverterFactory.create())

接口：

interface SubmitService {
@POST("one")
suspend fun submitCarton(@Body json: String): Response<myModel>
}

.addConverterFactory(ScalarsConverterFactory.create())

我自己解决了

.addConverterFactory(ScalarsConverterFactory.create())

问题是哑的，改型2给出了404，即使web服务返回错误消息

.addConverterFactory(ScalarsConverterFactory.create())

增加

implementation 'com.squareup.okhttp3:logging-interceptor:3.12.1'

private val interceptor = HttpLoggingInterceptor().setLevel(HttpLoggingInterceptor.Level.BODY)

private val okHttpClient = OkHttpClient().newBuilder()
    .connectTimeout(1, TimeUnit.MINUTES)
    .readTimeout(30, TimeUnit.SECONDS)
    .writeTimeout(20, TimeUnit.SECONDS)
    .addInterceptor(interceptor)
    .build()

.addConverterFactory(ScalarsConverterFactory.create())

发现改装发送了一个非常未格式化的字符串

"{ \"all my json filled with \" }" 

.addConverterFactory(ScalarsConverterFactory.create())

而不是

{ json } 

.addConverterFactory(ScalarsConverterFactory.create())

通过添加

.addConverterFactory(ScalarsConverterFactory.create())

到我的服务厂

.addConverterFactory(ScalarsConverterFactory.create())

如果有人想知道为什么我基本上将json创建为字符串而不是使用json对象，是因为我所谈论的服务真的希望它以非常特定的顺序排列，json只是不关心它，但它希望它看起来也像json

.addConverterFactory(ScalarsConverterFactory.create())