Error Using fscanf to open file
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what should i do to resolve this problem?
Error using fscanf
Invalid file identifier. Use fopen to generate a valid file identifier.
Error in data_proc1 (line 232)
a = fscanf(fid,'%e',inf); % write all data into a column of matrix a.
3 commentaires
Stephen23
le 10 Juil 2019
When calling fopen it is always a good idea to display the message if the file cannot be opened:
[fid,msg] = fopen(...);
assert(fid>=3,msg)
Please show us that complete message.
Guillaume
le 12 Juil 2019
Totally unrelated to the question, but since I saw that in your code, the whole:
if step < 10
filename_minute = ['00000',num2str(step)];
end
if step >=10 & step < 100
filename_minute = ['0000',num2str(step)];
end
if step >=100 & step < 1000
filename_minute = ['000',num2str(step)];
end
if step >=1000 & step < 10000
filename_minute = ['00',num2str(step)];
end
if step >=10000 & step < 100000
filename_minute = ['0',num2str(step)];
end
if step >=100000
filename_minute = [num2str(step)];
end
can be replaced by just:
filename_minute = num2str(step, '%06d');
Réponses (2)
Monika Phadnis
le 10 Juil 2019
The file you want to open should be in the same folder that is currently open in MATLAB for your script. You can browse to the folder where the script is. Is the file in the same folder?
Guillaume
le 10 Juil 2019
Why the conversion of filename to char? If it's not already a char array what is it?
You should always check the return value of fopen. Clearly, it failed to open the file. The most common reason is that your current directory is not where the file is. And the easiest way to fix that is to use absolute paths
folder = 'C:\somewhere\somefolder'; %use absolute path
[fid, errmsg] = fopen(fullfile(folder, char(filename))); %why the char?
assert(fid > 0, 'Failed to open file, error was: %s', errmsg);
Also, rather than clearing the a variable, which has a meaningless name, simply reuse your region variable, which is a much better name anyway:
region = fscanf(fid,'%g',inf);
fclose(fid);
region = reshape(region, nx, ny);
6 commentaires
Guillaume
le 12 Juil 2019
"I have trying ur suggestion, and this is the result: Failed to open file, error was: 78"
The code you've attached is not the one that produced that result.
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