Matlab 创建一个脚本，允许用户重复输入值，直到输入每个案例

我正在尝试创建一个脚本，允许用户重复输入x值，直到输入每个案例。我有三种可能的情况：

---

• 当

  x去除所有
  计数器
  变量时，输入案例1。改为使用
  boolean/logical
  标志，指示何时执行了特定案例。此外，您还需要检查每个案例本身是否已经执行了特定案例。不要将此作为单独的外部
  if
  语句来执行。这可能就是你最初写的那句话不起作用的原因。因此，做这样的事情。我将把
  %//NEW
  放在我添加新代码的地方：
  
    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
  
  
  我输入-1，尝试看看它是否拒绝负数，这是正确的。我输入了一个数字，它是
  <7
  ，也就是3。它成功地进入了第一个案例。我试着输入另一个数字，即
  <7
  ，即2。它给出的信息是，案例1已经运行。我试着输入一个介于7和12之间的数字。。。所以我试了8次。它相应地生成案例2。我用10再试一次，它说案例2已经运行了。最后，我尝试14，也就是
  >12
  ，它生成Case#3并在所有Case成功运行时停止
  
  我相信这就是你想要的
  
    if counter1>=1 || counter2>=1 || counter3>=1
        disp('That case has been run already');
  
    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
  
  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!