Why fopen does not accept the path when it is assigned to a variable.

3 visualizaciones (últimos 30 días)
Bilge
Bilge el 7 de Jun. de 2013
filename=['tests/testbkc/result-1/abc.txt']; fid=fopen('filename','w')
Here fid returns bigger than 3 but it does not create any file.
If I wrote the path directly into fopen then everything is okey. But I need to use variable.
what should be the type of filename? Isnt it char?
Does anybody have an idea? Thanks

Iniciar sesión para responder a esta pregunta.

Respuesta aceptada

Ahmed
Ahmed el 7 de Jun. de 2013
omit the the '' to get the content of the variable filename.
fid=fopen(filename,'w')

Más respuestas (0)

Categorías

Más información sobre Filename Construction en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by