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Dialogflow v2 API + Actions v2 API:必须设置MalformedResponse'finimum_response'

我正在尝试与Dialgoflow v2 API一起开始使用Google Actions v2 API.

我有以下示例(到目前为止在Dialogflow - > Fulfillment Webhook中)取自官方的Google Actions Migration Guide,但不幸的是我不断收到MalformedResponse'final_response'必须设置错误.

'use strict';

const functions = require('firebase-functions');
const { dialogflow } = require('actions-on-google');

const app = dialogflow();

app.intent('Default Welcome Intent', conv => {
  conv.ask('How are you?');
});

exports.dialogflowFirebaseFulfillment = functions.https.onRequest(app);
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响应是:

{
  "responseMetadata": {
    "status": {
      "code": 13,
      "message": "Failed to parse Dialogflow response into AppResponse because of empty speech response",
      "details": [
        {
          "@type": "type.googleapis.com/google.protobuf.Value",
          "value": "{\"id\":\"542fe4a8-6017-429f-81c3-61ba568e3659\",\"timestamp\":\"2018-04-19T20:16:25.606Z\",\"lang\":\"en-us\",\"result\":{},\"status\":{\"code\":200,\"errorType\":\"success\"},\"sessionId\":\"1524168985362\"}"
        }
      ]
    }
  }
}
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请知道为什么会发生这种情况?

actions-on-google dialogflow-es

7
推荐指数
1
解决办法
2139
查看次数

Haskell/ghci异常:bind:资源繁忙(地址已在使用中)

我正在尝试下面的代码(代码是这里的副本).问题是,当我用ctrl + c关闭服务器并尝试再次运行时,我得到:*异常:bind:资源繁忙(地址已在使用中).

在listenOn的文档中写道:注意:为了避免在GHC-Users邮件列表上多次弹出"已使用的地址"问题,我们在侦听套接字上设置了ReuseAddrsocket选项.如果您不想要此行为,请使用较低级别的听取.请问我该如何解决这个问题?(ghci版本7.6.3)

import Network (listenOn, accept, PortID(..))  
import Network.Socket (Socket, isSupportedSocketOption, SocketOption(..))
import System.IO (hSetBuffering, hGetLine, hPutStrLn, BufferMode(..), Handle)                                                                        
import Control.Concurrent (forkIO)                                        

echoImpl :: Handle -> IO ()                                               
echoImpl client = do                                                      
    line <- hGetLine client                                                
    hPutStrLn client line                                                  
    echoImpl client                                                        

clientHandler :: Socket -> IO ()                                          
clientHandler socket = do                                                        
    (client, _, _) <- accept socket                                          
    hSetBuffering client NoBuffering                                       
    forkIO $ echoImpl client                                               
    clientHandler socket                                                     

felix :: IO () 
felix = do
    let …
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networking haskell ghci

6
推荐指数
0
解决办法
1075
查看次数

Apache Spark:错误执行器 -&gt; 迭代器

我正在尝试从这里运行示例(Learning Spark book),但出现以下错误:

16/10/07 01:15:26 ERROR Executor: Exception in task 0.0 in stage 0.0 (TID 0)
java.lang.AbstractMethodError: com.oreilly.learningsparkexamples.mini.java.WordCount$1.call(Ljava/lang/Object;)Ljava/util/Iterator;
    at org.apache.spark.api.java.JavaRDDLike$$anonfun$fn$1$1.apply(JavaRDDLike.scala:124)
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我开始工作: $SPARK_HOME/bin/spark-submit --class com.oreilly.learningsparkexamples.mini.java.WordCount ./target/learning-spark-mini-example-0.0.1.jar ./README.md ./wordcounts

请想法,为什么会发生这种情况?

完整日志:

mini-complete-example$ $SPARK_HOME/bin/spark-submit --class com.oreilly.learningsparkexamples.mini.java.WordCount ./target/learning-spark-mini-example-0.0.1.jar ./README.md ./wordcounts
Using Spark's default log4j profile: org/apache/spark/log4j-defaults.properties
16/10/07 01:15:23 INFO SparkContext: Running Spark version 2.0.0
16/10/07 01:15:23 WARN NativeCodeLoader: Unable to load native-hadoop library for your platform... using builtin-java classes where applicable
16/10/07 01:15:24 INFO SecurityManager: Changing view acls to: eDS
16/10/07 01:15:24 …
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apache-spark

3
推荐指数
1
解决办法
3212
查看次数

haskell:将两个数组合并为一个Data数组

你能跟我帮忙吗?

我有2个数组,我需要合并到一个数据数组.下面的代码进行合并,但它创建了一个笛卡儿...

我想跟随输出:

DataXY {var1 = "x1", var2 = "y1"},
DataXY {var1 = "x2", var2 = "y2"},
DataXY {var1 = "x3", var2 = "y3"}
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- -码

data  DataXY = DataXY {
   var1 :: String,
   var2 :: String
} deriving (Eq, Show)

parse ::  [DataXY]
parse = x
    where
        x = [DataXY v1 v2 | v1 <- arr1, v2 <- arr2]
            where
                arr1 = ["x1", "x2", "x3"]
                arr2 = ["y1", "y2", "y3"]
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谢谢,m.

haskell

1
推荐指数
1
解决办法
437
查看次数