想从网站下载文件，json 文件来自 https://api.openrouteservice.org/ 但我收到此错误：java.net.MalformedURLException: no protocol: [Ljava.lang.String;@f2b1656
这是代码
private fun downloadUrl(strUrl: String): String{
var fullData = ""
try {
URL(strUrl).openStream().use { inp ->
BufferedInputStream(inp).use { bis ->
val data = ByteArray(1024)
var count: Int
while (bis.read(data, 0, 1024).also { count = it } != -1) {
fullData += data.toString()
}
}
}
}catch (e: Exception){
Log.d("DownloadTask", e.toString())
}
return fullData创建网址
private fun getDirectionURL(): String{
val destination: String = coordinates[2].toString() + "," + coordinates[3].toString()
val origin: String = globalLocation.longitude.toString() + "," + globalLocation.latitude.toString()
val urlString: String = java.net.URLEncoder.encode("https://api.openrouteservice.org/v2/directions/driving-car?api_key=5***&start=" + origin + "&end=" + destination, "UTF-8")
Log.d("urlCreate", urlString)
return urlString
}网址有效，我在浏览器中检查了它
回答1
您的问题出在 val urlString: String
您可以使用字符串插值来正确构建 URL，也可以使用 Android 方法使用 URI builder 和
附加路径（字符串）
追加查询参数（参数和 Value）。
https://developer.android.com/reference/android/net/Uri.Builder
回答2
java.net.URLEncoder.encode(yourUrl).toString() 将返回：
https%3A%2F%2Fapi.openrouteservice.org%2Fv2%2Fdirections%2Fdriving-car%3Fapi_key%3D5***%26start%3D2%2C3%26end%3D2%2C3java.net.MalformedURLException：无协议：https%3A% 2F%2Fapi.openrouteservice.org%2Fv2%2Fdirections%2Fdriving-car%3Fapi_key%3D5***%26start%3D2%2C3%26end%3D2%2C3
我不认为你需要用 URLEncoder 对象覆盖你的 url。只需返回您的网址，它就会起作用。
private fun getDirectionURL(): String{
val destination: String = coordinates[2].toString() + "," + coordinates[3].toString()
val origin: String = globalLocation.longitude.toString() + "," + globalLocation.latitude.toString()
val urlString: String = "https://api.openrouteservice.org/v2/directions/driving-car?api_key=5***&start=$origin&end=$destination"
Log.d("urlCreate", urlString)
return urlString
}