创建一个脚本，允许用户重复输入值，直到输入了每一种情况。

Question

我正在尝试创建一个脚本，允许用户重复输入x值，直到输入了每一种情况。我有三个可能的案例：

• 当x <= 7时输入案例1
• 在7 <= x <= 12时输入案例2。
• 在x > 12时输入案例3

我想使用一个while语句来出错--检查用户输入，确保x > 0。每次输入一个案例时，我都要打印案例编号&创建的y值：

• 案例1的y = x^3 + 3
• 案例2的y = (x-3)/2
• 案例3的y = 4*x+3

任何案件都不能进行两次审理。如果发生这种情况，脚本将输出类似于“已运行的情况”之类的内容。一旦所有案例都被输入，我想打印类似于‘所有案例都被输入’的东西。

以下是我到目前为止尝试过的：

  counter1 = 0;
  counter2 = 0;
  counter3 = 0;
  while counter1==0 || counter2==0 || counter3==0
  x = input('Please enter an x value > 0: ');
  while x < 0
      x = input('Invalid! Please enter another x value > 0: ');
  end
  if counter1>=1 || counter2>=1 || counter3>=1
      disp('That case has been run already');
  elseif x<=7
      counter1 = counter1 + 1;
      y = x.^3 + 3;
      fprintf('Case 1: y = %d \n',y);
  elseif 7<x && x<=12
      counter2 = counter2 + 1;
      y = (x-3)./2;
      fprintf('Case 2: y = %d \n',y);
  elseif x>12
      counter3 = counter3 + 1;
      y = 4.*x+3; 
      fprintf('Case 3: y = %d \n',y);
  else counter1==1 && counter2==1 && counter3==1;
  end
  end
  disp('All cases have been entered!')

我现在唯一不能工作的就是这部分：

  if counter1>=1 || counter2>=1 || counter3>=1
      disp('That case has been run already');

它似乎完全被忽视了。有什么建议吗？

Answer 1

去掉所有的counter变量。使用boolean / logical标志来指示何时执行了特定的情况。此外，您还需要检查每个案例本身是否已经执行了特定的案例。不要将其作为单独的外部if语句执行。这可能就是你最初写的那句话行不通的原因。就这样做这样的事。我将把%//NEW放在我添加了新代码的地方：

  case1 = false; %// NEW
  case2 = false; %// NEW
  case3 = false; %// NEW
  while ~case1 || ~case2 || ~case3 %// NEW: While at least one of the cases has not been run...
      x = input('Please enter an x value > 0: ');
      while x < 0
          x = input('Invalid! Please enter another x value > 0: ');
      end

      if x <= 7 %// NEW
         if case1 %// NEW: Check if case #1 has already been run
             %// If it has, show this to the user, then continue in the loop
             fprintf('Case #1 has already been run!\n');
             continue; %// NEW - Continue through the loop.  Don't do anything else
         end

         case1 = true; %// NEW - Set to true if we haven't run this case already
         y = x.^3 + 3;
         fprintf('Case 1: y = %d \n',y);

      elseif 7<x && x<=12
         if case2 %// NEW - Repeat like Case #1 here
            fprintf('Case #2 has already been run!\n'); %// NEW
            continue; %// NEW
         end

         case2 = true; %// NEW
         y = (x-3)./2;
         fprintf('Case 2: y = %d \n',y);

      elseif x>12
         if case3 %// NEW - Repeat like Case #3 here
             fprintf('Case #3 has already been run!\n'); %// NEW
             continue; %// NEW
         end

         case3 = true; %// NEW
         y = 4.*x+3; 
         fprintf('Case 3: y = %d \n',y);
       end %// End if
   end %// End while
   disp('All cases have been entered!') %// Display once all cases have been entered

下面是一个运行示例，说明它是有效的：

Please enter an x value > 0: -1
Invalid! Please enter another x value > 0: 3
Case 1: y = 30 
Please enter an x value > 0: 2
Case #1 has already been run!
Please enter an x value > 0: 8
Case 2: y = 2.500000e+00 
Please enter an x value > 0: 10
Case #2 has already been run!
Please enter an x value > 0: 14
Case 3: y = 59 
All cases have been entered!

我输入-1是为了看看它是否拒绝负数，这就是。我输入了一个数字< 7，是3，它成功地进入了第一个情况。我尝试输入另一个数字，即< 7，即2，它给出了这样的消息:第一种情况已经运行了。我试图输入一个介于7到12之间的数字.因此，我尝试8。它相应地生成案例#2。我用10再试一次，它说第二个案例已经运行了。最后，我尝试14，即> 12，它生成Case #3并停止，因为所有情况都已成功运行。

我相信这就是你要找的。