How to use 'for loop' to check user input?

I am trying to ask a user to input a value between 0-1. I then want to check that value against pre-existing variables r1b, r2b, r3b to see if it is greater and output the corresponding column as disp. So if user enters 0.15, it will be greater than r2b and r3b and the return will be columns r2a and r3a.

I have code to display the dialog box and return an error if not correct, but am not sure how to incorporate the loop?

r1a= column

r2a= colum

r3a= column

r1b=0.1

r2b=0.2

r3b=0.3

exit=false;

msg='Please enter an R^2 value for Cement:';

while ~exit

data = str2double( inputdlg(msg) );

exit = (0<= data1 && 1>=data1);

if ~exit

msg = 'Input must be between the values 0-1. Please re-enter: ';

end

end

Answer 1

[Nandini]

To check the user input against pre-existing variables and output the corresponding columns based on the condition, you can use a loop and conditional statements. Here's an example of how you can modify your code to achieve this:

r1a = column1;
r2a = column2;
r3a = column3;
r1b = 0.1;
r2b = 0.2;
r3b = 0.3;

exit = false;
msg = 'Please enter an R^2 value for Cement:';

while ~exit
    data = str2double(inputdlg(msg));
    exit = (0 <= data && 1 >= data);
    
    if ~exit
        msg = 'Input must be between the values 0-1. Please re-enter: ';
    else
        if data > r1b
            disp(r1a);
        end
        if data > r2b
            disp(r2a);
        end
        if data > r3b
            disp(r3a);
        end
    end
end

I hope this helps! Let me know if you have any further questions.

[Lauren-Xante Claassen]

Thank you Nandini!

I'm not sure I explained this as well as I could.

If the columns were from a Matrix (x), could I return the results in a new matrix?

x = 1 2 3

1 2 3

1 2 3

a =3

b= 6

c= 9

User input is 5.

loop statement

Return display:

y = 2 3

2 3

2 3

Answer 2

[VBBV]

modify the upper limit in exit condition line according to the user input,

x = [ 1    2    3
      1    2    3
      1    2    3];
r1a = x(:,1);
r2a = x(:,2);
r3a = x(:,3);
r1b = 3;
r2b = 6;
r3b = 9;

exit = false;
msg = 'Please enter an R^2 value for Cement:'

msg = 'Please enter an R^2 value for Cement:'

if ~exit
    data = str2double('5')  % // e.g. user input
    exit = (0 <= data && 10 >= data)  % modify the upper limit in  line according the user input
    
    if ~exit
        msg = 'Input must be between the values 0-1. Please re-enter: ';
    end
     if data > r1b & (data < r2b  | data < r3b)
            disp([r2a r3a])
     elseif data > r2b & (data < r1b  | data < r3b)        
            disp([r1a r3a])
     elseif data > r3b & (data < r1b  | data < r2b)                
            disp([r1a r2a])
     end    
end

data = 5

exit = logical
   1

     2     3
     2     3
     2     3

[VBBV]

you might want to add more conditions if the user input is less than all other values in conditions , in which case it need to prompt user for required output. Here i have used if-else conditional statement to demonstrate, since using a while loop it will never exit the loop, so its better to include a condition to exit the loop when the user inputs / outputs are all satisfied. Preferabl, a break statement with a condition inside the loop is necessary to exit the loop

x = [ 1    2    3
      1    2    3
      1    2    3];
r1a = x(:,1);
r2a = x(:,2);
r3a = x(:,3);
r1b = 0.1;
r2b = 0.2;
r3b = 0.3;

exit = false;
msg = 'Please enter an R^2 value for Cement:'

msg = 'Please enter an R^2 value for Cement:'

if ~exit
    data = str2double('0.15')  % // e.g. user input
    exit = (0 <= data & 1 >= data)  % modify the upper limit in  line according the user input
    
    if ~exit
        msg = 'Input must be between the values 0-1. Please re-enter: ';
    end
     if data > r1b & (data > r2b  & data > r3b)
            disp('Specify the output required')
     elseif data > r1b & (data < r2b  & data < r3b)
            disp([r2a r3a])
     elseif data > r2b & (data < r1b  & data < r3b)        
            disp([r1a r3a])
     elseif data > r3b & (data < r1b  & data < r2b)                
            disp([r1a r2a])
     end    
end

data = 0.1500

exit = logical
   1

     2     3
     2     3
     2     3