i know that if 1 is present at the 4th position of binary representation of attribute then this is a directory, but i am not sure if 1 is not present at that location should i consider it as a file?
or is there any other attribute present to determine folder or file ?
please help me.
Thanks.
Every file has a File Record in Master File Table (MFT) of the volume.
You can check the 2-byte flag stored at 0x16 and 0x17(attention, little endian). The second bit (counting from right) tells whether it's a folder(1), or a file(0).
if (flag & 0x02)
it's a folder
else
it's a file
If you change this bit that would originally represent a file to 1 by force, for example with the help of WinHex, and (probably a restart or system cache fresh is needed) double click it, OS would report that the file is corrupted.
In addition, the first bit tells if it is deleted.
if (flag & 0x01)
it's a normal file or folder not deleted
else
it's a deleted file or folder
Related
I have a list of MD5 hash of files stored in a text file. And I want delete them all when it been found on system or a path. But I have problem to code it. I have tried to but it only scan one file from listed MD5 so its not what i needed. Is there any way to find them and deleted files which there MD5 hash's are listed in a path. Thanks.
pidgin pseudocode:
put md5s in array
cycle through a filesystem
for each file, put into varable, compute md5hash of variable
if md5hash is in array, delete file
maybe you should skip swap files and system folders.
I have 500,000 files which I need to read in Fortran and each file has ~14,000 entries in it (each entry is only about 100 characters long). I need to process each line for each file at a time. For example, I need to process line 1 for all 500,000 files before moving on to line 2 from the files and so forth.
I cannot open them all at once (I tried making an array of file pointers and opening them all) because there will be too many files open at once. Instead, I would like to do something as follows:
do iline = 1,Nlines
do ifile = 1,Nfiles
! open the file
! read a line
! close the file
enddo
end
In hopes that this would allow me to read one line at a time (from each file) and then move on to the next line (in each file). Unfortunately, each time I open the file it starts me off at line 1 again. Is there any way to open/close a file and then open it again where you left off previously?
Thanks
Unfortunately it is not possible in this way in standard Fortran. Even If you specify
position="ASIS"
the actual position will be unspecified for a not already connected unit and will be in fact the beginning of the file on most systems.
That means You have to use
read(*,*)
enough times to get on the right place in the file.
You could also use stream access. The file would be again opened at the beginning, but you can use
read(u,*,pos=n) number
where n is the position saved from the previous open. You can get the position from
inquire(unit=u, pos=n)
n = n
You would open the file with acess="STREAM".
Also 500000 opened files is indeed too much. There are ways how to inquire for the system limits and how to control them, but also your compiler may have some limits http://www.cyberciti.biz/faq/linux-increase-the-maximum-number-of-open-files/
Other solution: Couldn't you store the content of the files in memory? Today couple of Gigabytes is OK, but it may be not enough for you.
You can try using fseek and ftell in something like the following.
! initialize an array of 0's
do iline = 1,Nlines
do ifile = 1,Nfiles
! open the file
! fseek(fd, array(ifile))
! read a line
! array(ifile)=ftell(fd)
! close the file
enddo
end
The (untested) idea is to store the offset of each file in an array and position the cursor at that place upon opening the file. Then, once a line is read, the ftell retrieves the current position which is saved to memory for next round. If all entries have the same length, you can spare the array and just store one value.
If the files have fixed, i.e., constant, record lengths, you could use direct access. Then you could "directly" read a specific record. A big "if" however.
the overhead of all the file opening/closing will be a big performance bottleneck.
You should try to read as much as you can for each open operation given whatever memory you have:
pseudocode:
loop until done:
loop over all files:
open
fseek !as in damiens answer
read N lines into array ! N=100 eg.
save ftell value for file
close
end file loop
loop over N output files:
open
write array data
close
I have 7 files with extensions like xyz.rar.001 - xyz.rar.007 clearly they are parts of a single file. I have all the 7 parts. I join them using a file joiner into a single file xyz.rar and try to unrar them with WINRAR , it says that archive is corrupted It is clear that 1 or 2 parts are corrupted. IS THERE ANY WAY TO FIND THEM ? Please help I don't want to re download all of them NOTE- winrar can detect a corrupt part if the parts were splitted using winrar (with extensions like part1.rar , part2.rar etc. ) but not if they are named as rar.001
Parts .001 - .006 should have the same size. Check if there is a file with a different byte size.
Are there multiple files in the RAR or just the one? With multiple you could run a Test and see which is the first file to fail.
I think it's strange that there is a second tool used to split the RAR archive up. (e.g. HJSplit) This lets me think that .002 could be a RAR archive too. Try opening xyz.rar.001 with WinRAR and test/exctract. It happens more that RAR archives have the extension .001 instead of .rar. An example.
Naming your archives in WinRAR like this can be accomplished by putting "xyz.rar.001" as Archive name on the General tab and checking "Old style volume names" on the Advanced tab.
If I then join the files with HJSplit, I get one .rar file (that is corrupt). When I Test it, it says "Next volume is required". In the diagnostic messages I can see "The required volume is absent" and "CRC failed in X. The file is corrupt"
If there is one file stored inside the RAR and the RAR is indeed just chopped up into 7 pieces, there is no way of telling without additional files such as .sfv or .par2. (unless the RAR does not use compression: you can parse the underlying file for errors and calculate the part where it goes wrong)
I ran into this problem when uploading a file with a super long name - my database field was only set to 50 characters. Since then, I have increased my database field length, but I'd like to have a way to check the length of the filename before uploading. Below is my code. The validation returns '85' as the character length. And it returns the same count for every different file I upload (none of which have a file name length of 85).
<cfscript>
missing_info = "<p>There was a slight problem with your submission. The following are required or invalid:</p><ul>";
// Check the length of the file name for our database field
if ( len(Form["ResumeFile1"]) gt 100 )
{
missing_info = missing_info & "<li>'Resume File 1' is invalid. Character length must be less than 100. Current count is " & len(Form["ResumeFile1"]) & ".</li>";
validation_error = true;
ResumeFileInvalidMarker = true;
}
</cfscript>
Anyone see anything wrong with this?
Thanks!
http://www.cfquickdocs.com/cf9/#cffile.upload
After you upload the file, the variable "clientFileName" will give you the name of the uploaded file, without a file extension.
The only way to read the filename before you upload it would be to use JavaScript to read and parse the value (file path) in the file field.
A quick clarification in the wording of your question. By the time your code executes the file upload has already happened. The file resides in a temporary directory on the ColdFusion server and the form field related to the file upload contains the temporary filename for that file. Aside from checking to see if a file has been specified, do not do anything directly with that file or you'll be circumventing some built in security.
You want to use the cffile tag with the upload action (or equivalent udf) to move the temp file into a folder of your choosing. At that point you get access to a structure containing lots of information. Usually I "upload" into a temporary directory for the application, which should be outside of the webroot for security.
At this point you'll then want to do any validation against the file, such as filename length, file type, file size, etc and delete the file if it fails any checks. If it passes all checks then you move it into it's final destination which may be inside the webroot.
In your case you'll want to check the cffile structure element clientFile which is the original filename including extension (which you'll need to check, since an extension doesn't need to be present and can be any length).
I have a folder of files with scrambled file names. The file extensions are scrambled too. The folder contains a variety of different file formats. The files are not encrypted.
example: original file name = abcde.pdf
scrambled file name = !##FDZ13
Is there a way to recover the original file names? If not, is there a way to differentiate the file formats (.pdf, .png, ...)? Ultimately, I wish to access and use these files again.
I am working with windows.
Wei, in principle, the case is quite easy.
I assume you know the set of file types that can possibly appear there. Let's say we expect there to be DOC, PDF and PNG files.
Then I would go ahead and do the following:
- create a subdirectory for every file type you expect
- for each file f
- for each file type t
- move f under a nice name with appropriate file extension
to the subdirectory for file type t
- try to open the file with the correct application for t
- continue with next file if it works
- otherwise continue with next file type
- at this point the directory should contain no files anymore
- move all files from the subdirectories back to this one
- remove the subdirectory.