I see no way to set destination directory or file here: http://www.cs.cmu.edu/~quake/triangle.switch.html
Actually, the program places result file in the same directory, even if current directory is different.
Why? Is ti possible to change?
The output files for the program are generated from the input file names. You can see this from the source code on line 3586
strcpy(b->outnodefilename, b->innodefilename);
...
strcat(b->outnodefilename, ".node");
strcat(b->outelefilename, ".ele");
...
Because of that I don't think there is a way to set the output directory as an option. It seems you will need to manually copy the output files to a different directory
cp output.node your/output/dir/output.node && rm output.node
Related
I have a simple script in matlab and I want to load a file. It seems it only works if the file is in the same dir as the script. If I add the file to a directory it does not read it.
For example:
fileID = fopen('myfile','r' ,'n', 'US-ASCII');
but when I put myfile in files:
fileID = fopen('files/myfile','r' ,'n', 'US-ASCII');
or
fileID = fopen('./files/myfile','r' ,'n', 'US-ASCII');
I get a -1 as a fileID. File cannot be read.
As per the comments, this is happening because you most likely added the path of where the script was located to your MATLAB path but you did not add the subdirectory where the file was in to your path. This is why it can't find the file. Therefore to avoid this in the future, you need to physically change the directory (i.e. the Working Directory) of where MATLAB is currently operating to where your script is stored.
It is then where local referencing should work. You can do this by either using the cd function, going to the top of your MATLAB window where you see the directory listing, clicking on the arrow to the right and pulling a drop down menu to change the directory, typing the actual directory you want by clicking on any blank space in the directory listing to enable a text box:
... or if you are running the code in the MATLAB editor, it'll request that you change directories as the script you are trying to run is not currently located in the working directory.
You can also programmatically add the subfolders in your script directory using mfilename, fileparts, genpath and addpath:
[dir, ~, ~] = fileparts(mfilename('fullpath')); % locate your script directory
addpath(genpath(fullfile(dir))); % add the folder and all subfolders to Matlab search directory
% then load your file.
fileID = fopen('myfile','r' ,'n', 'US-ASCII')
If it is also important that all outputs should be placed within the same directory as your script file, you can cd to your script directory:
cd(dir)
The compile command mcc -m app.m -a file.ext -a ./dir creates a directory /app containing app.m and file.ext and a directory /dir on the same level as /app containing all the files in /dir. What is the solution to add /dir in the /app directory, not on the same directory level (i.e. /app/dir)?
(Here is why I want to do this: In directory /dir are stored the images that are used by app.m, such as splash screen, button icons, default images, etc. app.m is accessing them using imread('./dir/img.jpg'). Since the compiler is adding the /dir directory one level below where it appears in the Matlab structure at development time, the images are no longer accessible when the standalone software is deployed. Hence I need to use a isdeployed switch to specify the correct path to the images for the development and deployment cases. I would rather avoid this, probably on code aesthetic grounds of an inconsistency of treating added files differently according to whether they are or not on the same directory level as the compiled application [file file.ext is put in /app while the image files from /dir are moved on the same level as /app].)
Instead of adding the path with the compile command, you may be able to use mkdir at run time.
Your command would look like this:
status=mkdir('dir');
The advantage to this is that the dir path is now below your app path, i.e. /app/dir.
The disadvantage is that the user of your program will need the privileges to create that directory.
If you need to compile the directory with the app, you could still use mkdir and then movefile to move all of your files to the new directory from the packaged one.
A possible solution is to use two paths depending on whether the application is deployed or not. Code example:
% path to images directory 'pix'
apppath = mfilename('fullpath');
idx = strfind(apppath,filesep);
if isdeployed
pixdir = [apppath(1:idx(end-1)),'pix',filesep];
else
pixdir = [apppath(1:idx(end)),'pix',filesep];
end
% read image
img = imread(fullfile(pixdir,'logo.jpg'));
example
There is a file "sample.rar".
Folder structure is: "rising\dawn\ and here there are many (folders1, folders2 and file1, file2)" in this archive.
i have used following command
7z.exe x "sample.rar" "rising\dawn\*" -oi:\delete
The result is:
all files and folders in "rising\dawn\" are extracted to "i:\delete" folder but the empty parent folders "rising\dawn\" are also created in destination folder.
e.g. destination looks:
i:\delete\rising\dawn\folder1\file1.bmp
i:\delete\rising\dawn\folder2\subfolder
i:\delete\rising\dawn\file1.txt
i:\delete\rising\dawn\file2.txt
i don't want "rising\dawn\" empty folders to be created but the folder structure there onwards must be as is in the archive.
i want the result:
i:\delete\folder1\file1.bmp
i:\delete\folder2\subfolder
i:\delete\file1.txt
i:\delete\file2.txt
at last i found a way out solution. thanks to the winrar support. i have accepted it as an answer below.
if you find the question useful don't forget to click the up-vote button.
Finally this gave me the result.
Thanks to winrar support.
rar x -ep1 sample.rar rising\dawn\* d:\e\delete\
i have tried other answers given here, this is the only correct answer.
don't forget to upvote.
You can extract the archive normally and
1) move the lower level folder/files to where you would like it, then
2) remove the extra top level archive folders.
Code to do so will depend on the exact task.
Using e command instead of x and add -r option works well.
Like this:
7z.exe e -r "sample.rar" "rising\dawn\*" -oi:\delete
My executable version is "7-Zip [64] 9.20 2010-11-18",
And the platform is Windows 8.1.
This command line eliminates unnecessary parent folders and preserves the hierarchy of folders.
You need to use the e command rather than the x command:
7z.exe e "sample.rar" "scholar\update\*" -oi:\delete
Using e instead of x means 7zip will extract all matching files into the same folder (as specified via the -so switch, or the current directory if this isn't specified) rather than preserving the folder structure from inside the archive.
So i had this issue that occurred when I ran a Matlab script. Here is an a simple example that illustrates it:
So its important to outline the file structure:
MainFolder
script.m
SubFolder
a1.csv
a2.csv
a3.csv
now say i have a script like this:
-> script.m
dir
it would simply print out the files in the folder.
Now the wierd thing, if i run the script in the Subfolder like this:
>>script
it will do this:
>> a1.csv a2.csv a3.csv
but if i do this in the folder:
>>run('C:\Users\....\MainFolder\script.m')
it will only print out
>> script.m
So obviously it is acting as if i ran it form MainFolder rather than SubFolder.
What is the point of this functionality?
The dir command shows the directory contents of Matlab's current directory, not that of where the script is located. So the script showed you the directory contents of wherever you happened to be in the Matlab command prompt when you called that script.
To get what you want, use this in the script:
dir(fileparts(mfilename('fullpath')))
Use pwd to see current dir
Use cd to change directory
Use path to see if your project folders are included in the path
Use which to see you are calling the right *.m file (in case there is multiple .m files with same name on the path)
I am using the system command in MATLAB as follows (with the current directory being 'scripts'):
[status, result] = system('cd ..\\TxtInOut')
However, invoking the system command does not seem to work. It returns status = 0 and result = ''.
Any suggestions?
If you want to change directories, you should use the CD command. The argument can be either a full path or relative path:
cd('c:\matlab\toolbox'); %# Full path to a directory
cd('scripts'); %# Move to a subdirectory "scripts"
cd('..\TxtInOut'); %# Move up one level, then to directory "TxtInOut"
If you want information about a directory, you should use the DIR command. DIR will return an m-by-1 structure of information for a directory, where m is the number of files and folders in the directory. Again, the argument can be either a full path or relative path:
data = dir('c:\matlab\toolbox'); %# Data for a full path to a directory
data = dir('scripts'); %# Data for a subdirectory "scripts"
NOTE: When working on different platforms (i.e. Windows or UNIX), you will have to pay attention to whether you use the file separator \ or /. You can get the file separator for your platform using the function FILESEP. You can also build your file paths using the function FULLFILE.
Any command executed by "system" is external to MATLAB. A command shell is generated, executes your request, and then returns the result. The 0 result indicates successful completion: the command shell changed its current directory as requested and then returned. (Command shells use non-zero to indicate an error, because there are usually many more ways that a program can fail than succeed.) Unfortunately that only affects the command shell's current directory - see gnovice's answer for how to actually change the directory.
you can use cd, dir, ls, etc directly in matlab without call system functions.
You can also use the underlying operating system commands by preceding them by an exclamation sign.
For instance:
!dir will show you the current directory contents in Windows
!pwd will show you the current directory in Linux/Mac
But calling cd does not change the current directory!