Using filepath input in fopen

8 次查看（过去 30 天）
Suha
Suha 2017-8-17
评论： Suha 2017-8-17
Hi all, I can't seem to use fopen to open a file using a variable that stores the name and path of the file. For example, the variable [pathvariable] stores /imper/codec/data.txt. I am doing fid = fopen([pathvariable],'r') but it is not working.
I've tried the following too:
fid = fopen(pathvariable,'r')
fid = fopen('pathvariable','r')
In all cases, I get a fid of -1.
Could someone please help. Thanks.
  3 个评论
显示 1更早的评论
Stephen23
Stephen23 2017-8-17
@Suha: in 99 percent of cases the filepath is incorrect or there is a spelling mistake somewhere. Check the filename carefully.
Suha
Suha 2017-8-17
Thank you KSSV and Stephen for your prompt reply ! My pathvariable was indeed the issue, which I have now fixed. :)

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Guillaume
Guillaume 2017-8-17
The proper syntax is indeed your first example:
fid = fopen(pathvariable, 'r');
There could be many reasons for why it does not work: invalid path(file not found), file locked, permission denied, etc. First thing to do would be to look at the errmsg output of:
[fid, errmsg] = fopen(pathvariable, 'r')
which should give you more info.
If you're on Windows, /imper/codec/data.txt does not look like a absolute path which may be the problem. I would always use absolute paths with fopen to avoid any issue:
fid = fopen(fullfile('C:\some\where', relativepath), 'r');
  1 个评论
Suha
Suha 2017-8-17
Thank you Guillaume for taking time out to write such a clear and detailed response. Much appreciated. Using the errmsg output I was able to spot that my pathvariable was incorrect, hence the file could not be detected. All fixed and working now. Many thanks !!

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Entering Commands 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by