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Creating a script that will allow the user to input values repeatedly until each case has been entered

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Jen
Jen 2014-11-19
回答： Image Analyst 2014-11-19
I'm trying to create a script that will allow the user to input x-values repeatedly until each case has been entered. I have three possible cases: Case 1 is entered when x<=7, case 2 is entered when 712. I want to use a while statement to error-check the user input, ensuring that x>0. Each time a case is entered, I want to print the case number & the created y-value (y = x^3 + 3 for case 1, y = (x-3)/2 for case 2, and y = 4*x+3 for case 3). No case may be ran twice (then the script will output something like 'That case has been run already'). Once all cases have been entered, I want to print something like 'All cases have been entered'.
Here's the script I have so far & I'm stuck:
x = input('Please enter an x value > 0: ');
while x < 0
x = input('Invalid! Please enter another x value > 0: ');
end
counter1 = 0;
counter2 = 0;
counter3 = 0;
if 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);
end
if counter1==0 || counter2==0 || counter3==0
x = input('Please enter an x value > 0: ');
elseif counter1>=1 || counter2>=1 || counter3>=1
disp('That case has been run already')
elseif counter1==1 && counter2==1 && counter3==1
disp('All cases have been entered!')
end

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采纳的回答

Image Analyst
Image Analyst 2014-11-19
Try this:
clc;
counter1 = false;
counter2 = false;
counter3 = false;
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 x<=7
% It's case 1
if counter1
fprintf('Case 1 has been run already.\n');
continue;
end
counter1 = true;
y = x.^3 + 3;
fprintf('Case 1: y = %d \n',y);
elseif 7<x && x<=12
% It's case 1
if counter2
fprintf('Case 2 has been run already.\n');
continue;
end
counter2 = true;
y = (x-3)./2;
fprintf('Case 2: y = %d \n',y);
elseif x>12
% It's case 3
if counter3
fprintf('Case 3 has been run already.\n');
continue;
end
counter3 = true;
y = 4.*x+3;
fprintf('Case 3: y = %d \n',y);
end
end
fprintf('All cases have now been entered!\n')

更多回答（1 个）

Adam
Adam 2014-11-19
编辑：Adam 2014-11-19
You haven't said what you are stuck with, but a few points that immediately come to mind:
  • You are only looping on user input if the user enters a number < 0, otherwise you only ask them once and then once again at the end. You need a loop covering the whole thing, but with an exit condition.
  • Your second case should have an && not a || otherwise the final elseif will never be reached.
  • You shouldn't need an elseif for the final case, a simple else should do the job since it is the only case remaining.
  • You should do your counter checking before handling the cases I assume, otherwise you still do the maths anyway then tell the user the case was already run afterwards.
  • Your final elseif clause should form the exit condition from the all-enclosing while loop I mentioned above if you want the user to keep entering values until that clause is true.
  3 个评论
显示 1更早的评论
Guillaume
Guillaume 2014-11-19
In any of your elseif, the x doesn't change, so of course the next time you go through the elseif the same is selected since x is still the same value.
Somewhere in your while loop, you need to change x.
Jen
Jen 2014-11-19
Here's my revised code again:
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!')
Only thing I can't seem to get to work now is this part:
if counter1>=1 || counter2>=1 || counter3>=1
disp('That case has been run already');
It seems to be ignored entirely. Any suggestions?

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