I am creating a Zip file up to 4GB and using FileInstall() but I'm not able to extract and also not able to create a single .exe file.
FileInstall("myPath\myfile.zip","DestinationFolderName")
Is there a limitation regarding size for the FileInstall() function?
FileInstall() is not going to unzip your zip archive no matter what the size is. It will only place the zip in the destination. It's only used to include files in the compiled AutoIt script.
Also, if you use a folder name as the destination, you need a trailing backslash (\).
I highly recommend reading the entry for FileInstall() in the help docs.
Related
I'm providing a .zip with a .R file and a .xlsx file to some people
I need to make a code that can read this .xlsx file in any directory of any pc.
But as the directories vary from computer to computer, I couldn't find a solution.
IMPORTANT: I'm not using Rstudio for read this .R, so i just can use base functions
Using R - How do I search for a file/folder on all drives (hard drives as well as USB drives) This question don't solve my problem..
Take a look at the here package. When you load the library (library("here")) it sets "base" working directory and then you can use the package to construct relative file paths given that location. For example, if inside your .zip file you have an R script (e.g., My Data Analysis.R) that analyzes data that is kept within a folder called data you could read it in using, for example, read.csv(here("data", "my_csv_file.csv")) and it will construct the full appropriate file path no matter what computer it is on. Of course the file structure of the program needs to stay the same across programs.
I have an unarchiver that takes in an archive name, and a directory name, and dumps all files from that archive into that directory. No other command-line options. However, someone zipped a file in the archive I am looking to decompress, with 500-ish characters in the filename, and now that program fails when it hits that file (practically all file systems have a limit of 256). What alternative do I have, short of changing the source code and recompiling the unarchiver?
I must mount something as a directory, which would take the files that the unarchiver is writing, and dump them elsewhere-- possibly even as one big file. This something should not send fail messages, even if some write really did fail. Is this possible?
I'm new to R and frankly the amount of documentation is overwhelming, and I haven't been able to find the answer to this question.
I have created a number of .R script files, all stored in a folder that I can access on my server (let's say the folder is, using the Windows backslash character \\servername\Paige\myscripts)
I know that in R you can call each script individually, for example (using the forward slash required in R)
source(file="//servername/Paige/myscripts/con_mdb.r")
and now this script, con_mdb, is available for use.
If I want to make all the scripts in this folder available at startup, how do I do this?
Briefly:
Use your ~/.Rprofile in the directory found via Sys.getenv("HOME") (or if that fails, in R's own Rprofile.site)
Loop over the contents of the directory via dir() or list.files().
Source each file.
as eg via this one liner
sapply(dir("//servername/Paige/myscripts/", "*.r"), source)
but the real story is that you should not do this. Create a package instead, and load that. Bazillion other questions here on how to build a package. Research it -- it is worth it.
Far the best way is to create a package! But as first step, you could also create one r script file (collection.r) in your script directory which includes all the scripts in a relative manner.
In your separate project scripts you can than include only that script with
source(file="//servername/Paige/myscripts/collection.r", chdir = TRUE)
which changes the directory before sourcing. Therefore you would have only to include one file for each project.
In the collection file you could use a loop over all files (except collection.r) or simply list them all.
I backed up a large number of files to S3 from a PC before switching to a Mac several months ago. Several months later, I'm now trying to open the files and realized the files were all compressed by the S3 GUI tool I used so I can not open them.
I can't remember what program I used to upload the files and standard decompression commands from the command line are not working e.g.,
unzip
bunzip2
tar -zxvf
How can I determine what the compression type is of the file? Alternatively, what other decompression techniques can I try?
PS - I know the files are not corrupted because I tested downloading and opening them back when I originally uploaded to S3.
You can use Universal Extractor (open source) to determine compression types.
Here is a link: http://legroom.net/software/uniextract/
The little downside is that it looks in the first place for the extension, but I manage to change the extensions myself for a inknown file and it works almost always, eg .rar or .exe etc..
EDIT:
I found a huge list of archive programs, maybe one of them will work? It's ridiciously big:
http://www.maximumcompression.com/data/summary_mf.php
http://www.maximumcompression.com/index.html
I have a batchfile with SFTP instruction to download txt files (cron job), basically: get *.txt
Wondering what the best method is to delete those files after the server has downloaded them. The only problem being that the directory is constantly being updated with new files, so running rm *.txt afterwards won't work.
I've thought of a couple complex ways of doing this, but no command line based methods. So, I thought I'd shoot a query out to you guys, see if there's something I haven't thought of yet.
I suggest to make a list of all the files that were downloaded and then issue ftp delete/mdelete commands with the exact file names.