Hello everyone,
Some months ago I deployed a number of apps using the matlab compiler. I used to add a folder "images" within the "files required to the application to run" and successfully access it by doing:
fullfile(cftroot,'images')
After updating to 2019a, it seems that the folder hierarchy of the app is changed, and the images folder is located in a subfolder
fileName = 'fileName'; % first 12 characters of the name given in the matlab compiler
fullfile(cftroot, fileName, 'images')
My question is simple (I hope):
is there a function that allows me to automatically retrieve "fileName"?
(So that if I change the name of the deployed file, I don't need to change the code)
Thank you very much,
Rob

 Respuesta aceptada

Subhadeep Koley
Subhadeep Koley el 23 de Ag. de 2019

1 voto

MATLAB provides fileparts and mfilename function to get parts of file name and to get file name of currently running code.
Therefore, using the following code in the startupFcn callback of your code might help.
[filepath,name,ext] = fileparts(mfilename('fullpath'));
Here the variable "name" contains the name of the current running file.

3 comentarios

Roberto Gentile
Roberto Gentile el 25 de Ag. de 2019
Thank you!
very much appreciated!
Jiri Hajek
Jiri Hajek el 22 de Jul. de 2021
I'm looking for an answer to the same question, but this one is not really correct. An app name (exe file) may be different from the m-file before compilation. The mfilename function return the name of the original m-file, not of the exe-file.
Any help would be much appreciated...
Anna Weeks
Anna Weeks el 3 de Nov. de 2023
filepath = fileparts(mfilename('fullpath'));
Appname = strtok(filepath(numel(ctfroot)+1:end),'\');

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el 20 de Ag. de 2019

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el 3 de Nov. de 2023

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