这就是我想要做的
package gc.services.http.playWs.lmi
import com.softwaremill.macwire.Macwire
import play.api.Play.current
import play.api.libs.ws._
import play.api.libs.ws.ning.NingAsyncHttpClientConfigBuilder
import scala.concurrent.Future
import play.api.libs.ws.{WSResponse, WSRequestHolder, WSClient}
import scala.concurrent.Future
class LmiWs extends Macwire{
val baseUrl = "http://api.lmiforall.org.uk/api/v1"
//inject plays ws client
lazy val http = wire[WSClient]
def getRequestHolder(url: String = baseUrl): WSRequestHolder = {
http.url(url)
}
def get(request: WSRequestHolder): Future[WSResponse] = {
request.get()
}
}如果我尝试,wireWS IntelliJ给出的不能解析符号。在上面的IntelliJ中没有显示错误,但是在编译mac连线抛出时
WeakTypeTagplay.api.libs.ws.WSClient的非查找构造函数
这个类只是为我想要点击的应用程序定义基本路由,它将由不同的类对每个特定的路由进行扩展。我希望能够模拟WS依赖项,这样当我测试这个类和后代类时,我可以这样做,而无需发送实际的HTTP请求。
发布于 2015-10-05 04:17:24
也许你可以试试这个:
lazy val http: WSClient = WS.clienthttps://stackoverflow.com/questions/27877535
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