problems using && with dropdown list result
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hi!
S={'NH4Cl'; 'MgCl2'; 'CaCl2'; 'ZnCl2'; 'ZnSO4'; 'NaCl'};
>> str={'Choose the combinations of salts available'};
>> result=listdlg('Promptstring',str, 'ListSize', [100,100], 'ListString', S, 'SelectionMode', 'multiple');
>> if result==1&&result==3
disp('hei')
end
the error I get: Operands to the || and && operators must be convertible to logical scalar values. Donnt understand because the output from result is scalar and is should be possible to use && togehter with scalar outputs..
4 件のコメント
You explictly set selection mode to 'multiple' so unless you only select one item it won't be a scalar. And if it was a scalar then
result==1&&result==3
would clearly always evaluate to false anyway! Maybe you want something more like:
all( ismember( [1,3], result ) )
Muazma Ali
2019 年 9 月 24 日
編集済み: Muazma Ali
2019 年 9 月 24 日
the cyclist
2019 年 9 月 24 日
Muazma,
The problem with your code is clear to me (and Adam). You have allowed the user to select multiple items. For example, they might select 'NH4Cl' and 'CaCl2'. In this case,
result = [1 3]
and you are trying to evaluate
[1 3]==1 && [1 3]==3
which is
[1 0] && [0 1] % these are logical vectors
These are not allowable operands for the short-circuit logical operators. The operands must each be a single scalar value.
What is not clear to me is what you want to happen in your if statement. Perhaps you could explain this more completely.
Muazma Ali
2019 年 9 月 24 日
採用された回答
その他の回答 (1 件)
the cyclist
2019 年 9 月 24 日
My best guess as to what you want is
if any(ismember(result,[1 3]))
disp('hei')
end
The if statement will be entered if the user's selections include 'NH4Cl' or 'CaCl2' (or both).
2 件のコメント
Muazma Ali
2019 年 9 月 24 日
Guillaume
2019 年 9 月 24 日
Maybe the OP wants all instead of any. I.e all the selections must be NH4Cl or CaCl2
if all(ismember(result, [1 3]))
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