我是JVM和Scala以及Play 2.0的新手。
我正在将一个需要通过Authorize.net进行付款处理的遗留应用程序转换为Play。查看java.net.URL源代码,有很多潜在的失败点。在下面给出的接口中,您会在哪里实现try/catch块?我需要相应地调整方法签名,可能会返回Either [Error,Success]到调用客户端代码。
import java.net.{URL, URLEncoder}
import java.io.{BufferedReader, DataOutputStream, InputStreamReader}
import javax.net.ssl._
trait Authnet {
private val prodUrl = "https://secure.authorize.net/gateway/transact.dll"
private val testUrl = "https://test.authorize.net/gateway/transact.dll"
protected def authNetProcess(params: Map[String,String]) = {
val(conn, urlParams) = connect(params)
val request = new DataOutputStream( conn.getOutputStream )
request.write(urlParams.getBytes)
request.flush()
request.close()
val response = new BufferedReader(new InputStreamReader(conn.getInputStream))
val results = response.readLine().split("\\|")
response.close()
results.toList
}
private def connect(params: Map[String,String]) = {
val urlParams = (config ++ params) map { case(k,v) =>
URLEncoder.encode(k, "UTF-8") + "=" + URLEncoder.encode(v, "UTF-8")
} mkString("&")
lazy val url = if (isDev) new URL(testUrl) else new URL(prodUrl)
val conn = url.openConnection
conn.setDoOutput(true)
conn.setUseCaches(false)
(conn, urlParams)
}
private val config = Map(
'x_login -> "...",
'x_tran_key -> "...",
...
)
}