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matlab编写的gui程序已打包成xx.exe(gui控件的背景图片与xx.exe同目录D:\test)，双击xx.exe能正常运行。但将某种文件(如.txt)的默认打开方式设置为xx.exe后，双击.txt文件(注:.txt与xx.exe非同目录哦)
，发现xx.exe将找不到背景图片(系统的当前路径变成了.txt文件的路径)；求解决啊
本人希望双击.txt文件后xx.exe能获取xx.exe的路径,望高手指点，不胜感激！
以下是本人在测试时的一些发现：
1.双击xx.exe，xx.exe里边的目录及文件会被复制到系统临时文件夹中运行(C:\Users\...\AppData\Local\Temp\...\mcrCache8.0\)。
2.若将背景图片zz.jpg与.txt同目录，则xx.exe能找到zz.jpg；(图片的读取路径为imread('zz.jpg'))。
3.若用获取当前.m文件路径的方法，在双击.txt的情况下获取到的路径是C:\Users\...\AppData\Local\Temp\...\mcrCache8.0\...而不是D：test
4.若用 cd 或pwd，在双击.txt的情况下获取到的路径是C:\Windows\System32
5.若用path，在双击zz.txt的情况下获取到的路径是C:\Users\...\AppData\Local\Temp\...\mcrCache8.0\...
为什么在双击zz.txt后xx.exe不能找到自己(xx.exe)的路径呢，我又必须要用xx.exe目录下的一些图片，该怎么解决呢 纠结了好几天了
以下附上测试代码：
function test(fpath)
%方法1：
currentpath=mfilename('fullpath');
index=findstr(currentpath,'\');
currentpath=currentpath(1:index(end));
msgbox(currentpath);
%方法2：
currentpath=[cd,'\'];
msgbox(currentpath);
%方法3：
currentpath=[pwd,'\'];
msgbox(currentpath);
%方法4：
currentpath=path;
index=findstr(currentpath,';');
currentpath=currentpath(1:index(1)-1);
currentpath=[currentpath,'\'];
msgbox(currentpath);
try
if(exist(fpath))
msgbox('通过.txt打开');
end
catch
msgbox('直接运行的.exe');
end
m文件打包.exe的方法论坛里很多，这里就不写了
