在 if else 语句中发出

Question

我需要帮助来解决我遇到的这个问题。我已经解决了 Whosebug 中可用的几个问题，但我仍然不明白。

这是我正在写的代码：

var myFeedback = ""
      if (this.userFriendlinessRating <= 4.0 && this.userFriendlinessRating >= 3.5){
        myFeedback = 'VERY SATISFIED' 
        return myFeedback;
      } else if (this.userFriendlinessRating <= 3.5 && this.userFriendlinessRating >= 2.5){
        myFeedback = 'SATISFIED' 
        return myFeedback;
      } else if (this.userFriendlinessRating <=2.5 && this.userFriendlinessRating >=1.5){
        myFeedback = 'NOT SATISFIED' 
        return myFeedback;
      } else {
        myFeedback = 'VERY NOT SATISFIED' 
        return myFeedback;
      }
      console.log(myFeedback)

这是我得到的错误：

Answer 1

每个if部分和最后else部分都包含return，所以控件不能通过if-elseif-elseif-else结构。因此最后 console.Log 行无法访问。

删除所有内部 return 语句并在末尾添加一个以获得您想要的：

var myFeedback = ""
      if (this.userFriendlinessRating <= 4.0 && this.userFriendlinessRating >= 3.5){
        myFeedback = 'VERY SATISFIED' 
      } else if (this.userFriendlinessRating <= 3.5 && this.userFriendlinessRating >= 2.5){
        myFeedback = 'SATISFIED' 
      } else if (this.userFriendlinessRating <=2.5 && this.userFriendlinessRating >=1.5){
        myFeedback = 'NOT SATISFIED' 
      } else {
        myFeedback = 'VERY NOT SATISFIED' 
      }
      console.log(myFeedback)
      return myFeedback;

Answer 2

如果你想访问控制台日志，你应该像这样重组，所以最后一行是可以访问的。

var myFeedback = ""
if (this.userFriendlinessRating <= 4.0 && this.userFriendlinessRating >= 3.5){
  myFeedback = 'VERY SATISFIED' 
} else if (this.userFriendlinessRating <= 3.5 && this.userFriendlinessRating >= 2.5){
  myFeedback = 'SATISFIED' 
} else if (this.userFriendlinessRating <=2.5 && this.userFriendlinessRating >=1.5){
  myFeedback = 'NOT SATISFIED' 
} else {
  myFeedback = 'VERY NOT SATISFIED' 
}
console.log(myFeedback)
return myFeedback;

Answer 3

您必须为 myFeedback 变量的决定传递一个 return，并且 return 应该放在末尾。请参阅下面的示例大纲

      var obj = ""

      if (cond){
        obj = 'VERY SATISFIED' 
      } else if (cond){
        obj = 'SATISFIED' 
      } else if (cond){
        obj = 'NOT SATISFIED' 
      } else {
        obj = 'VERY NOT SATISFIED' 
      }
      console.log(obj)
      return obj;

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