Matlab unable to format string to path

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Sajjid Wajid
Sajjid Wajid 2022-1-6
编辑： Stephen23 2022-1-6
In original script there is part of code where is this
filepathsource='C:\Users\...';
strFP=dir([filepathsource '*.txt']);
Which works nicely. Problem is that i need to change that filepathsource to some more dynamic so i did.
filepathsource=string(strcat(app.folderPath,'\',string(app.currentCount),'\'));
strFP=dir([filepathsource '
*.txt']);
But i end with error
Error using dir
Name must be a text scalar.
And i need original formating because i get errors later in script. Is there any way how can i fix it? I tried everything but always is something somewhere wrong. And i just dont understand whats difference between original filepathsource and the one that i created bot are strings. (output of my filepathsource is correct when i display it)
  1 个评论
Stephen23
Stephen23 2022-1-6
编辑：Stephen23 2022-1-6
" Is there any way how can i fix it? "
Yes: use FULLFILE instead of concatenating strings and separators.

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回答（1 个）

Walter Roberson
Walter Roberson 2022-1-6
Use fullfile() instead of concatenation of characters or strings.
Your most immediate problem is that the results of
['characters' "string"]
is not 'charactersstring' or "charactersstring" and is instead ["characters" "string"] -- that is, [] between a character vector and a string is done by converting the character vector into a string object and then you have [] of two string objects which gives you a string array.
To join a character vector and a string object use strjoin() or use the + operator such as 'hello' + "there"
But all this is best avoided by using fullfile()

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