要将歌曲插入 echoprint 服务器数据库,您需要做的就是调用该ingest
方法。基本上,它只是一个带有正确 json 正文的 HTTP POST 请求。这是我正在使用的 Scala 代码(Java 将非常相似):
import EasyJSON.JSON
import EasyJSON.ScalaJSON
import dispatch.Defaults.executor
import dispatch._
class EchoprintAPI {
val API_URL = "http://your.api.server"
def queryURL(code: String) = url(s"$API_URL/query?fp_code=$code")
def query(code: String): scala.concurrent.Future[ScalaJSON] = {
jsonResponse(queryURL(code))
}
def ingest(json: ScalaJSON, trackId: String): scala.concurrent.Future[ScalaJSON] = {
val metadata = json("metadata")
val request = url(s"$API_URL/ingest").POST
.addParameter("fp_code", json("code").toString)
.addParameter("artist", metadata("artist").toString)
.addParameter("release", metadata("release").toString)
.addParameter("track", metadata("title").toString)
.addParameter("codever", metadata("version").toString)
.addParameter("length", metadata("duration").toString)
.addParameter("genre", metadata("genre").toString)
.addParameter("bitrate", metadata("bitrate").toString)
.addParameter("source", metadata("filename").toString)
.addParameter("track_id", trackId)
.addParameter("sample_rate", metadata("sample_rate").toString)
jsonResponse(request)
}
def delete(trackId: String): scala.concurrent.Future[ScalaJSON] = {
jsonResponse(url(s"$API_URL/query?track_id=$trackId").DELETE)
}
protected def jsonResponse(request: dispatch.Req): scala.concurrent.Future[EasyJSON.ScalaJSON] = {
val response = Http(request OK as.String)
for (c <- response) yield JSON.parseJSON(c)
}
}
要生成指纹代码,您可以使用echoprint-codegen
命令行调用或使用Java JNI 与 C lib 集成