scala 值 toInt 不是 Any 的成员
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scala value toInt is not a member of Any
提问by user644745
The println in the following code works (with or without toInt)
以下代码中的 println 有效（有或没有toInt）
println("retweets : ", e.getOrElse("retweets", 0).toInt)
top10Tweets(""+e.get("text").get, e.getOrElse("retweets", 0).toInt)
But when I pass it as an argument of a function (as above), it does not work. It says "value toInt is not a member of Any"
但是当我将它作为函数的参数传递时（如上），它不起作用。它说“值 toInt 不是 Any 的成员”
When I remove toInt, it says,
当我删除 toInt 时，它说，
type mismatch;
[error] found : Any
[error] required: Int
e is a Map, as follows,
e是一个Map，如下
def tweetDetails(obj: twitter4j.Status) = {
Map(
"id" -> obj.getUser().getId(),
"screenName" -> obj.getUser().getScreenName(),
"text" -> obj.getText(),
"retweets" -> obj.getRetweetCount(),
"mentions" -> obj.getUserMentionEntities().length)
}
signature of top10Tweets,
前10条推文的签名，
def top10Tweets(tweets: String, retweet_c: Int, mention_c: Int) = {
}
回答by drexin
edit:
编辑：
Ok, with the new information I would suggest you to create a case class that holds the data instead of using a Map, this way you will preserve type information. I know it is common to use hashes/maps for that in dynamically typed languages, but in statically typed languages as scala data types are the preferred way.
好的，有了新信息，我建议您创建一个保存数据的案例类，而不是使用 a Map，这样您就可以保留类型信息。我知道在动态类型语言中使用哈希/映射是很常见的，但在静态类型语言中，scala 数据类型是首选方式。
orig:
来源：
As I neither know what eis, nor what signature top10Tweetshas, I can only assume. But from your code and the error I assume that eis a Map[String, String]and you are trying to get the string representation of an integer for the key "retweets"and convert it to an Int. As a default value you pass in an Int, so the type inferencer infers type Any, because that is the most common super type of Stringand Int. However Anydoes not have a toIntmethod and thus you get the error.
因为我既不知道是什么e，也不知道签名top10Tweets有什么，我只能假设。但是根据您的代码和错误，我假设e是 aMap[String, String]并且您正在尝试获取键的整数的字符串表示形式"retweets"并将其转换为Int. 作为默认值传入 an Int，因此类型推断器会推断 type Any，因为这是Stringand最常见的超类型Int。但是Any没有toInt方法，因此您会收到错误消息。
Map("x" -> "2").getOrElse("x", 4).toInt
<console>:8: error: value toInt is not a member of Any
Map("x" -> "2").getOrElse("x", 4).toInt
Either pass in the default value as String, or convert the value of "retweets"to an Intbefore, if it exists:
无论是传递默认值String，或值转换"retweets"成Int之前，如果存在的话：
e.get("retweets").map(_.toInt).getOrElse(0)
Anyway a little more information would help to give an accurate answer.
无论如何，多一点信息将有助于给出准确的答案。
回答by daaatz
Yes because in Map is "string" -> "string" and You did when getOrElse ( else ) string -> int, thats why its Any.
是的，因为在 Map 中是“字符串”->“字符串”，而当 getOrElse（else）字符串-> int 时，您这样做了，这就是为什么它的 Any。
Map("x" -> 2).getOrElse("x", 4).toInt
works fine or You can:
工作正常，或者您可以：
Map("x" -> "2").getOrElse("x", "4").toInt
回答by Y.fei
The same problem bothers me before you could check below
在你查看下面之前，同样的问题困扰着我
Map("x" -> 2).getOrElse[Int](x, 4)
