Is there a way to construct a filepath that links to the Documents folder of the active user. So instead of C:\Users\User\Documents\ something like C:\Active_User\Documents\
ps. I try to make use of this in KNIME.
The file chooser elements in KNIME understand a URL in the form of "knime://knime.workflow" which accesses the current workflow location regardless of higher directory path.
You could also use a Java Variable Edit to get the username in Java, which you use to create a string that can be used by the File Reader (or other node) as a flow variable.
It depends on what you're trying to achieve.
You can use
C:\Users\%USERNAME%\Documents
which will use the environment variable %USERNAME% (= current user).
In C#/.NET you can use Environment.SpecialFolder.MyDocuments like this:
Console.WriteLine("GetFolderPath: {0}",
Environment.GetFolderPath(Environment.SpecialFolder.MyDocuments));
In Java System.getProperty("user.home"); should give you the right base diretory to start with.
Related
TLDR
I need to get paths to system directories like "Screenshots":
On an English system. I can just use this one:
C:/Users/User/Pictures/Screenshots
How do I get the path to "Screenshots" directory on a non-English system?
C:/Users/User/Pictures/[NAME]
Description
I have a file manager app, it displays system directories and loads them on click.
The app can run system commands via Powershell and use Node.js (preferred)
Problem
The problem is, it only works if the system has English system language.
Currently, to resolve the "Screenshots" directory path, the app simply joins the User directory with the word "Screenshots"
const pictures = electronRemote.app.getPath('pictures')
const screenshots = PATH.join(pictures, 'Screenshots')
link to the line in code
Expectedly, the C:/Users/User/Screenshots path only exists on English systems.
One way to solve this is to use short names, at least on Windows, I know that system directories have short names like SCREEN~1 and WALLPA~1 for Screenshots and Wallpapers directories, but if I use these names the paths will look like this:
C:/Users/User/SCREEN~1 instead of C:/Users/User/Screenshots throughout the app.
And even if I were to run these paths through a function to convert it to readable name, how would I know which word to replace it with? I need to get the name in the system's language.
Are these translations stored somewhere on the system? Can I just retrieve the translated directory name and use that in the code above?
Question
How do I make it to get / resolve the actual path of system directories like Screenshots and Wallpapers, independently of system locale?
If you know how to do it, could you please suggest the solution for all platforms (Win, Mac, Linux)?
Should I just use the short names like SCREEN~1 and then automatically replace all the occurrences in UI and also filter all paths through a function that replaces this short name with the actual path throughout the whole app? Seems like a lot of work, this approach
I am using the latest version of pyRevit, v45.
I'm writing some info in temporary files with
myTempFile = script.get_instance_data_file("id")
This creates a file named pyRevit_2018_xxxx_id.tmp in which I store useful info. If I'm not mistaken, the "xxxx" part is changing every time I reload Revit. Now, I need to get access to this information from another pyRevit script.
How can I retrieve the name of the temp file I need to read? In other words, how do I access "myTempFile" from within the second script, which has no idea of the name of "myTempFile"?
I guess I can share somehow that variable between my script, but what's the proper way to do this? I know this must be a very basic programming question, but I'm indeed not a programmer ;)
Thanks a lot,
Arnaud.
Ok, I realise now that my variables in the 1st script cease to exist after its execution.
So for now I wrote the file name in another file, of which I know the name.. That works.
But if there's a cleaner way to do this, I'd be glad to learn ;)
Arnaud
pyrevit.script module provides 4 different methods for creating temporary files based on their use case:
get_instance_data_file:
for data files marked with Revit instance pid. This means that scripts running on another instance will not see this temp file.
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_instance_data_file
get_universal_data_file:
for temp files accessible to all Revit instances and versions
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_universal_data_file
get_data_file:
Base method to get a standard temp file for current revit version
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_data_file
get_document_data_file:
temp file marked with active document (so scripts working on another document will not see this)
http://pyrevit.readthedocs.io/en/latest/pyrevit/script.html#pyrevit.script.get_document_data_file
Each method uses a pattern to create the temp file name. So as long as the call to the method is the same of different scripts, the method generates the same file name.
Example:
Script 1:
from pyrevit import script
tfile = script.get_data_file('mydata')
Script 2:
from pyrevit import script
tempfile = script.get_data_file('mydata')
In this example tempfile = tfile since the file id is the same.
There is documentation on each so make sure you take a look at those and pick the flavor that serves your purpose.
I have a Java Web App running on Tomcat on which I'm supposed to exploit Path traversal vulnerability. There is a section (in the App) at which I can upload a .zip file, which gets extracted in the server's /tmp directory. The content of the .zip file is not being checked, so basically I could put anything in it. I tried putting a .jsp file in it and it extracts perfectly. My problem is that I don't know how to reach this file as a "normal" user from browser. I tried entering ../../../tmp/somepage.jsp in the address bar, but Tomcat just strips the ../ and gives me http://localhost:8080/tmp/ resource not available.
Ideal would be if I could somehow encode ../ in the path of somepage.jsp so that it gets extracted in the web riot directory of the Web App. Is this possible? Are there maybe any escape sequences that would translate to ../ after extracting?
Any ideas would be highly appreciated.
Note: This is a school project in a Security course where I'm supposed to locate vulnerabilities and correct them. Not trying to harm anyone...
Sorry about the downvotes. Security is very important, and should be taught.
Do you pass in the file name to be used?
The check that the server does is probably something something like If location starts with "/tmp" then allow it. So what you want to do is pass `/tmp/../home/webapp/"?
Another idea would be to see if you could craft a zip file that would result in the contents being moved up - like if you set "../" in the filename inside the zip, what would happen? You might need to manually modify things if your zip tools don't allow it.
To protect against this kind of vulnerability you are looking for something like this:
String somedirectory = "c:/fixed_directory/";
String file = request.getParameter("file");
if(file.indexOf(".")>-1)
{
//if it contains a ., disallow
out.print("stop trying to hack");
return;
}
else
{
//load specified file and print to screen
loadfile(somedirectory+file+".txt");
///.....
}
If you just were to pass the variable "file" to your loadfile function without checking, then someone could make a link to load any file they want. See https://www.owasp.org/index.php/Path_Traversal
In my linux python app for Fedora I want to save user's work to /home/user/Documents/MyCoolApp/TheirGreatWork.txt
But I am not sure how to find the "Documents" folder if the user is not using English as their default language.
What is the right way to determine the right path so that files go in their "Documents" folder.
EDIT
Here is a which comes up if you change locales... showing how paths can get easily changed.
I'd use the subprocess module to get the output of the command xdg-user-dir DOCUMENTS. For example:
import subprocess
documents_dir = subprocess.check_output(["xdg-user-dir", "DOCUMENTS"])
print documents_dir # This is what you're looking for.
There is no right way as the user may have changed their locale, which (fortunately) does not rename the directory. If you want a fixed place for files managed by your app, use ~user/.MyCoolApp or let the user specify the directory.
How can I get the current user path in Linux? It can be either with the GTK+ framework APIs, or plain C++.
Assuming you mean the current directory of the process:
The plain POSIX C function is getcwd().
In glib, there's also g_get_current_dir().
If you want to get home directory use getenv("HOME")
g_get_home_dir() from Glib is more cross-platform than getenv("HOME"). It also prefers /etc/passwd entries over the HOME variable for various reasons discussed at the aforementioned link.
Not sure whether you're wanting the contents of $PATH or the user's current working directory. However to cover both options...
PATH is an environment variable, so you can access this with getenv(), in this instance getenv("PATH"), and is defined in <stdlib.h>.
The current working directory can be obtained with getcwd(), and is defined in <unistd.h>.