how to access zip files in the directory?

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hi ..example.if i have a folder name as 'DATA',inside this subfolder namely A TO Z are ther..in this each folder(A TO Z),many .zip files are ther..i need to fetch all the .zip files inside main folder('DATA'),and extract them to a separate folder.is it possible? any defined function?

Accepted Answer

Laurent
Laurent on 5 Sep 2013
Using the 'dir'-command you can obtain a list of files.
In your case you could use something like:
filelist=dir('DATA/*.zip');
or
filelist=dir('DATA\*.zip');
depending on your OS.
Then filelist will be a struct in which filelist.name has the filenames.
'unzip' can then be used to unzip your files.
  10 Comments
sandy
sandy on 16 Sep 2013
Edited: sandy on 16 Sep 2013
this is the code i checked for my case,its not working
folder ='F:\test folder\';
for kk=2000:2013
if exist(folder,'dir')
for ll=1:12
if exist(folder,'dir')
for mm=1:31
currfold=[folder(mm) '\*.zip'];
currfiles=dir(currfold);
for jj=1:length(currfiles)
unzip([folder(ii) '\' currfiles(jj).name],outputfolder);
end
end
end
end
end
end
Laurent
Laurent on 16 Sep 2013
You are always using 'folder' and are not adding the year, month or date, when you check if they exist. I would change it in something like this:
Be aware that I did not check this code, so there could be small errors/mistakes, but I hope you get the idea.
for kk=2000:2013
yearfolder=[folder num2str(kk) '\'];
if exist(yearfolder,'dir')
for ll=1:12
monthfolder=[yearfolder num2str(ll) '\'];
if exist(monthfolder,'dir')
for mm=1:31
currfold=[monthfolder num2str(mm) '\'];
if exist(currfold,'dir')
currfiles=dir([currfold '*.zip']);
for jj=1:length(currfiles)
unzip([currfold '\' currfiles(jj).name],outputfolder);
end
end
end
end
end
end
end

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