Each test case saves results to a separate UITestActionLog.html file. But in the end of each test case I'd like to move that .html to a different folder and rename it.
Is it possible to do so in, say, [TestCleanup()]? If yes, then how can I programmatically get .html report location?
The TestContext class contains several fields with "directory" in their names. These can be used to access the various directories associated with running the tests.
As well as managing the files as asked by your question the TestContext class has an AddResultFile method. The Microsoft documentation on this mehod is not clear, but it seems that the files are saved for failing tests and discarded for passing tests.
To get the directory in which the UITestActionLog file will be located, use the TestContext.TestResultsDirectory Property. You can use below code to get the full path:
string fullPath = TestContext.TestResultsDirectory +"\" +"UITestActionLog.html";
Related
In the piece of code I have, there are many instances where I have the following line
'/home/myname/directory'
For example, I have the following lines of code
filepath = os.listdir('/home/myname/directory')
for content in filepath
# do something
In the next part of the project, I have to share the code with some one else. I know this person runs openSUSE. If I want code to create that specific directory with the same path to the directory as mine, what do I need to include?
I know its going to involve the OS module but i am not sure which functions and methods to use.
Your code can check the existence of the directory and create it if not found:
if not os.path.exists("/home/myname/directory"):
os.makedirs("/home/myname/directory")
# do something
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.
Does anyone know how to get the full path of a DirectoryEntry object in a Chrome Packaged App, without any tildes or other shortcuts?
I am writing a Google Chrome Packaged App. My app has a button where a user can choose a directory using chrome.fileSystem API. When the directory choice comes back to my app, it is represented by a DirectoryEntry object, which is defined in the File API. The object looks like this in the console:
DirectoryEntry {
filesystem: DOMFileSystem
fullPath: "/to_read"
isDirectory: true
isFile: false
name: "to_read"
__proto__: DirectoryEntry
}
I am using Windows and the full path of the directory is
C:\Users\David\Desktop\to_read
I would like a function that can return that path or something close. Unfortunately, the closest thing I found is chrome.fileSystem.getDisplayPath, but that returns the following:
~\Desktop\to_read
The return value from getDisplayPath is not useful to me, because I want to get the full name of the directory (including the drive) so I can compare it to some other full directory paths I have.
I tried calling toURL() on the DirectoryEntry and it returned an empty string.
A bit about my project: I want to write an iTunes library synchronizer as a Chrome Packaged App. The iTunes library XML file contains full paths like file://localhost/D:/David/Music/Bob/Bob%20Album/01%20Bob.mp3. The user will give my app access to his music folders, and I want to be able to tell if he gave me access to the right folders.
The only full paths available are those returned by getDisplayPath.
The mediaGalleries API may be a better fit for your project: http://developer.chrome.com/apps/mediaGalleries.html#iTunes.
If you have a full path path/to/the/file.mp3 and you want to load the file directly with tis you can do it but the solution is not perfect. Use the chrome.mediaGalleries API to ask where the user saves his music (in your case), then you can write a loop who check if one of the path repository is equal to a gallery.
For example if you got a file path/to/the/file.mp3 from the xml file and your galleries list looks like ["D", "to", "Videos"], write a function to check if each component of the file's path is a gallery. In this case your code will find "to", so you can launch a second function who use "the/file.mp3".
The second function has to use the given path and find if the gallery contains the right folders and finally the right file (use this example by Google). In the case you're trying to find "the/file.mp3" with the gallery to your loop has to find a directory named "the" then "file.mp3" (write a recursive function), if you find the file open it, otherwise come back to the first function if you haven't check all the galleries or all the path's component.
This is currently (2014-01-10) not a feature of Chrome, but I have suggested it and they are working on it:
https://code.google.com/p/chromium/issues/detail?id=322952
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
I'm trying to wrap code that requires two *.db4o data files for easy use. I've added the data files to my eclipse .classpath by placing the files in ${project_dir}/res/ and adding the line:
<classpathentry kind="src" path="res"/>
to my .classpath.
I then defined a default constructor to my wrapper class that takes no arguments but goes and finds the paths to the *.db4o files (the paths are required by the compiled code I'm using to set things up). My approach for getting the paths is:
String datapath = ClassLoader.getSystemResource("resource_name").getPath();
This works great when I debug/run my code in eclipse. However when I export it as a jar, I can see that the *.db4o files are in the jar, as well as my compiled code, but the path returned to "datapath" is of the form:
datapath = ${pwd}/file:${absolute_path_to_jar}!/{resource_name}
Is there something about the resource being inside of the jar that prevents an absolute path from working? Also, why is the behavior different simply because the code and resources live in a jar file? One last note is that while my application is intended for wider use (from PIG, python, etc. code) I'm testing it from Matlab which is where I'm getting the odd value assigned to "datapath".
Thanks in advance for any responses.
getSystemResource() returns URL to resource. If your resource is zipped in a jar file then the URL will point into it (with the "!" notation). getPath() returns the "path" part of the URL, not always an actual file path. URL can be one of many things, not just a file.