如果/ Else / Switch返回错误的结果

3 回答

• 2

  switch 语句的正确用法是

  switch (this.state.playerOnePick) {
      case 'Rock':
          switch (this.state.playerTwoPick) {
              case 'Scissors'):
              case 'Lizard':
                  return "Player One wins!";
                  break; // unnecessary after `return` but well
              case 'Paper':
              case 'Spock':
                  return "Player Two wins!";
                  break; // as above
          }
          break;
      case 'Lizard':
          switch (this.state.playerTwoPick) {
              case 'Spock':
              case 'Paper':
                  return "Player One wins!"
              case 'Scissors':
              case 'Rock':
                  return "Player Two wins!";
          }
          break;
  }
  

  你所展示的是 if / else 的布局，有很多布尔条件：

  if (this.state.playerOnePick === 'Rock') {
      if ((this.state.playerTwoPick === 'Scissors') || (this.state.playerTwoPick === 'Lizard')) {
          return "Player One wins!";
      } else if ((this.state.playerTwoPick === 'Paper') || (this.state.playerTwoPick === 'Spock')) {
          return "Player Two wins!";
      }
  } else if (this.state.playerOnePick === 'Lizard') {
      if ((this.state.playerTwoPick === 'Spock') || (this.state.playerTwoPick === 'Paper')) {
          return "Player One wins!";
      } else if ((this.state.playerTwoPick === 'Scissors') || (this.state.playerTwoPick === 'Rock')) {
          return "Player Two wins!";
      }
  }
  

  然而，实施Rock-Paper-Scissors-Lizard-Spock的真正问题就是重复（这给错误留下了很大空间） . 实际的编程任务是弄清楚如何减少它 .
  提示：为每个可能的选择分配一个整数，并使用一些数学 .
  写一个单独的函数 winner(pick1, pick2) ，在第一个玩家获胜时返回 -1 ，在领带时返回 0 ，在第二个玩家获胜时返回 1 . 然后简单地从与UI内容有关的ReactJS代码中调用它 .

  回复于 2024-04-26T07:30:01+08:00
• 0

  开关盒不起作用 . 如果你想使用switch case，你可以这样做：

  switch(playerOnePick) {
   case 'Rock':
     switch(playerTwoPick) {
      case 'Scissors':
      case 'Lizard':
        // Player 1 wins
        break;
      case 'Paper':
      case 'Spock':
      // Player 2 wins
      break;
     }
     break;
   case 'Lizard':
     switch(playerTwoPick) {
      case 'Spock':
      case 'Paper': 
        // Player 1 wins
       break;
      case 'Scissors':
      case 'Rock': 
      // Player 2 wins
      break;
   }
  }
  

  等等

  编辑：您编写的代码是为if语句构造的，如果用if替换switch和case，它将起作用 .

  回复于 2024-04-26T07:30:01+08:00
• 0

  因为你已经这样设定了它 . 看最后 <h1> .

  说，你可以写一个更好的 switch/case 语句来评估这个条件 . 你高度重复自己，这是反对DRY（不要重复自己）的校长 .

  您可以将 <h2>P1: {this.state.playerOnePick} P2: {this.state.playerTwoPick}</h2> 完全拉出 switch 块并在开关关闭后立即放置 } . 或者你最好每个案例只放一个单一的值，而不是在一个案例中合并它们 . 这使得遵循代码变得更加困难 . 这是我对这个特殊开关盒的选择：

  switch (this.state.playerOnePick === 'Rock') {
      case(this.state.playerTwoPick === 'Scissors'):
           return ( <div> <h1>Player One wins !</h1> </div>);
           break;
      case(this.state.playerTwoPick === 'Lizard')):
           return ( <div> <h1>Player One wins !</h1> </div>);
           break;
      case(this.state.playerTwoPick === 'Paper'):
           return ( <div> <h1>Player Two wins !</h1> </div>);          
           break;
      case(this.state.playerTwoPick === 'Spock'):
           return ( <div> <h1>Player Twi wins !</h1> </div>);        
           break;
      }
  // The same goes for second switch
  <div><h2>P1: {this.state.playerOnePick} P2:{this.state.playerTwoPick}</h2><div>
  

  回复于 2024-04-26T07:30:01+08:00