"if "function
显示 更早的评论
how to write "IF" function to only accept numbers NO letters
need to ask input number (if user type any letters need to ask again to give number)
i know basic IF function but i need to know how to verified the input as number or letter ?
thank you Vijay
采纳的回答
Jiro Doke
2012-2-19
You can use the isstrprop function. If answer is the input you got from your user (using the input command),
all(isstrprop(answer, 'digit'))
Here's an example:
while true
answer = input('Enter a number: ', 's');
if all(isstrprop(answer, 'digit'))
answer = str2double(answer);
break;
else
disp('Only numbers supported');
end
end
Here's another way. This way is slightly better because it allows decimal points:
while true
answer = input('Enter a number: ', 's');
answer = str2double(answer);
if ~isnan(answer)
break;
else
disp('Only numbers supported');
end
end
2 个评论
vijayasinthujan vijayaratnam
2012-2-19
Jan
2012-2-19
The first method isstrprop(x,'digit') rejects the decimal point also, while the 2nd method accepts inputs as "3.14". In addition "1e3" should be accepted also, such that I prefer the STR2DOUBLE method.
更多回答（3 个）
Image Analyst
2012-2-19
Did you try using ischar() and isnumeric()? Or try something like this, or similar:
% Ask user for a number.
defaultValue = 45;
userPrompt = 'Enter the integer';
caUserInput = inputdlg(userPrompt, 'Enter the numeric value',1,{num2str(defaultValue)});
integerValue = round(str2num(cell2mat(caUserInput)));
% Check for a valid integer.
if isempty(integerValue)
% They didn't enter a number.
% They entered a character, symbols, or something else not allowed.
integerValue = defaultValue;
message = sprintf('I said it had to be an integer.\nI will use %d and continue.', integerValue);
uiwait(warndlg(message));
end
3 个评论
vijayasinthujan vijayaratnam
2012-2-19
Image Analyst
2012-2-19
They are built in functions to check if the argument is a character or a number. Are they not in your help?
Jan
2012-2-19
Simply use "help ischar" and "help isnumeric".
Walter Roberson
2012-2-19
0 个投票
In MATLAB, "if" is not a function: it is a control statement.
We've been nagging MathWorks for years to create a function equivalent to "if", but with no success yet :(
类别
在 帮助中心 和 File Exchange 中查找有关 Data Type Identification 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
