how to access zip files in the directory?

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?

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Laurent
Laurent el 5 de Sept. de 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 comentarios

sandy
sandy el 5 de Sept. de 2013
thanks...here, the .zip files are present inside the folders from A TO Z,not inside DATA itself.. i need to access all the zip file inside them(A TO Z). it like accessing files from all the subfolders(A to Z) of main folder(DATA)
Laurent
Laurent el 5 de Sept. de 2013
Editada: Laurent el 5 de Sept. de 2013
Ah, something like this could then be your solution:
folders=('A':'Z')';
outputfolder='yourfolder';
for ii=1:26
currfold=[folders(ii) '\*.zip'];
currfiles=dir(currfold);
%now do something with these files
for jj=1:length(currfiles)
unzip([folders(ii) '\' currfiles(jj).name],outputfolder);
end
end
sandy
sandy el 5 de Sept. de 2013
thanks..it works for A TO Z...but my another question...for example...i am giving the path of folder(YEAR).inside this folder.month folder are present as 01 TO 12,inside each month folder ,date folders are present from 01 to 31,inside each days folder alone zip files present.ineed to access this and unzip..nw its possible??
Yes this is definitely possible. You could use a similar approach for the years and months and put it into extra for-loops around the one from my example.
for kk=1997:2013
{code}
for ll=1:12
{code}
for mm=1:31
{my code}
end
end
end
Of course if not all dates/months/years are there you will have to include a way to check for this.
Also please accept my answer if it was useful for you.
sandy
sandy el 5 de Sept. de 2013
Editada: sandy el 5 de Sept. de 2013
thanks...its works..but if the any day or month folder is not present,what conditions can i use? any predefined function or code ther?
To check if a specific folder exists you can use:
if exist(folder,'dir')
{do what you need}
end
sandy
sandy el 16 de Sept. de 2013
thanks...i checked with your code..its not checking the existing files and not unzipping my files..any other idea?
Laurent
Laurent el 16 de Sept. de 2013
Difficult to say. Could you provide the exact code that you used and the exact location of these files? And did you get any errors?
sandy
sandy el 16 de Sept. de 2013
Editada: sandy el 16 de Sept. de 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
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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