How do I check if a folder exists which has a variable name?
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Hi,
I want to check if the created folder already exists. But the name of the folder depends on the file I work with.
fname = Measurement_20190829_18_15_10.bin
if ~exist('Figures' fname(11:end-4), 'dir')
mkdir('Figures', fname(11:end-4));
end
I have no clue how to make this work.
I always get Invalid expression with a marker on fname.
Please help.
Thank you!
3 commentaires
Splitting the extension from a filename using indexing is very fragile and a bad way to write code. The simple recommended method is to use fileparts:
[~,name,ext] = fileparts(...)
Does this really run without error?:
fname = Measurement_20190829_18_15_10.bin
What do you expect this syntax to do?:
exist('Figures' fname(11:end-4), 'dir')
% ^^^^^^^^^^^^^^^^^^^^^^^^^ what is this supposed to do?
When I try it, it throws an error.
David Müller
le 29 Août 2019
"How does fileparts help me in this example?"'
Instead of using this fragile, very non-robust code to obtain the filename without the extension:
fname(11:end-4)
you can simply use robust fileparts:
[~,name] = fileparts(fname);
"But the exist does not work properly."
exist does work properly. Read the last four lines of my previous comment.
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Johannes Fischer
le 29 Août 2019
Modifié(e) : Johannes Fischer
le 29 Août 2019
Your code is trying to store in fname the content of the field 'bin' of the struct 'Measurement_20190829_18_15_10', which propably doesnt exist.
fname needs to be a char array:
fname = 'Measurement_20190829_18_15_10.bin';
I have no idea about your folder structure, but Im guessing its 'Figures\t_20190829_18_15_10' ? If so, you would need to write
mkdir(['Figures\', fname(11:end-4)]);
and the same as the argument for 'exist'
5 commentaires
David Müller
le 29 Août 2019
"But the exist does not work properly."
Exist works properly. Your error is because of this:
'Figures'fname(11:end-4)
Note how you simply place two character vectors next to each other, without concatenting them or as inputs to a function. You should use fullfile on them.
David Müller
le 29 Août 2019
Stephen23
le 29 Août 2019
"..but how should it look like with my code."
See my answer.
It uses more robust fileparts rather than fragile indexing.
It uses simple fullfile to join the folder and file names.
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