In the terminal, is there a way to create a directory with a file in it in one step?
Currently I do this in 2 steps:
1. mkdir foo
2. touch foo/bar.txt
Apparently, touch foo/bar.txt doesn't work.
With only standard unix tools, the most direct way to create a directory and a file in this directory is
mkdir foo && touch foo/bar.txt
Unix is built around the philosophy of simple, single-purpose tools with the shell as a glue to combine them. So to create a directory and a file, you instruct a shell to run the directory creation utility then the file creation utility.
I won't swear that there isn't some bizarre way of using a standard tool that lets you do it with a single command. (In fact, there is: unpack an archive — except that you'll need to provide that archive as a file, with predefined owner, date and other metadata, or else use another command to build an archive.) But whatever it is would be convoluted.
Related
I want to version control my R scripts so I've created an R project and a GitHub repo. My scripts are scattered through several directories within the same directory where the R project is.
I would like that my GitHub repository harbors only the scripts, independently of the folders they are locally stored in. However when I run the below command:
git add folder/file.R
git commit -m "my_message"
git push -u origin master
A directory named folder is created containing file.R but I'd like to just see file.R without the folder. Do you know how can I do this? Also, would it be good practice? My local folders are organized so each directory contains its own scripts and results, that's the reason the scripts are separated.
Thank you very much
is there a way to add the file.R without specifying the path?
Not using git add, no. The design constraint for git add is that it should store the file's name exactly as it appears, including the forward slashes, so if the file's name is folder/file.R, that's the file's name.
You have some options here though:
You can make a parallel directory where you put the files with the names you want them to have. Run git init in that directory, copy the folder/file.R file to file.R in that directory. Then cd ../gitdir or whatever is appropriate to get there, and git add file.R.
This method is probably the best because it's the simplest.
You can write your own programs using git hash-file -w and git update-index, which are two of Git's plumbing commands. A plumbing command, in Git, is basically a command that exists so that you can build user-facing commands: they're not meant to be run by humans but rather by other programs. So you write a program (in whatever language you like) that uses these plumbing programs to achieve whatever you want.
In particular, you can create or find a Git blob object holding the contents of file.R as read from anywhere you like, then use git update-index to create an index entry holding whatever path you like and referring to the blob object you created (or found) with git hash-object with the -w flag.
Since Git is a suite of tools, not a solution, you can come up with your own method. The tools in Git are made with particular approaches in mind, but they are flexible enough to be repurposed.
Using any of the standard Robot libraries, is it possible to recursively copy the contents of a directory to an existing destination directory?
Basically, I'm looking for the equivalent of the following shell command: cp -r foo/. bar (note the trailing dot)
I tried Copy Directory but this creates a directory foo inside bar (as documented) and it doesn't stop doing that even when supplying the trailing dot. Copy Files chokes when it encounters a directory.
Is there anything I overlooked? Or do I need to just call cp -r myself?
As I only need this to work on Linux, I ended up implementing a custom keyword calling cp -r. If this is ever needed cross-platform, then I'll follow the suggestions to directly implement it in Python.
Copy Directory Contents
[Documentation] Recursively copies the contents of the source directory into the destination.
[Arguments] ${source} ${destination}
Directory Should Exist ${source}
Directory Should Exist ${destination}
${result} = Run Process cp -r ${source}/. ${destination}/
Should Be Equal As Integers ${result.rc} 0
I am using Robot Framework, to run 50 Testcases. Everytime its creating following three files as expected:
c:\users\<user>\appdata\local\output.xml
c:\users\<user>\appdata\local\log.html
c:\users\<user>\appdata\local\report.html
But when I run same robot file, these files will be removed and New log files will be created.
I want to keep all previous run logs to refer in future. Log files should be saved in a folder with a time-stamp value in that.
NOTE: I am running robot file from command prompt (pybot test.robot). NOT from RIDE.
Could any one guide me on this?
Using the built-in features of robot
The robot framework user guide has a section titled Timestamping output files which describes how to do this.
From the documentation:
All output files listed in this section can be automatically timestamped with the option --timestampoutputs (-T). When this option is used, a timestamp in the format YYYYMMDD-hhmmss is placed between the extension and the base name of each file. The example below would, for example, create such output files as output-20080604-163225.xml and mylog-20080604-163225.html:
robot --timestampoutputs --log mylog.html --report NONE tests.robot
To specify a folder, this too is documented in the user guide, in the section Output Directory, under Different Output Files:
...The default output directory is the directory where the execution is started from, but it can be altered with the --outputdir (-d) option. The path set with this option is, again, relative to the execution directory, but can naturally be given also as an absolute path...
Using a helper script
You can write a script (in python, bash, powershell, etc) that performs two duties:
launches pybot with all the options you wan
renames the output files
You then just use this helper script instead of calling pybot directly.
I'm having trouble working out how to create a timestamped directory at the end of the execution. This is my script it timestamps the files, but I don't really want that, just the default file names inside a timestamped directory after each execution?
CALL "C:\Python27\Scripts\robot.bat" --variable BROWSER:IE --outputdir C:\robot\ --timestampoutputs --name "Robot Execution" Tests\test1.robot
You may use the directory creation for output files using the timestamp, like I explain in RIDE FAQ
This would be in your case:
-d ./%date:~-4,4%%date:~-10,2%%date:~-7,2%
User can update the default output folder of the robot framework in the pycharm IDE by updating the value for the key "OutputDir" in the Settings.py file present in the folder mentioned below.
..ProjectDirectory\venv\Lib\site-packages\robot\conf\settings.py
Update the 'outputdir' key value in the cli_opts dictionary to "str(os.getcwd()) + "//Results//Report" + datetime.datetime.now().strftime("%d%b%Y_%H%M%S")" of class _BaseSettings(object):
_cli_opts = {
# Update the abspath('.') to the required folder path.
# 'OutputDir' : ('outputdir', abspath('.')),
'OutputDir' : ('outputdir', str(os.getcwd()) + "//Results//Report_" + datetime.datetime.now().strftime("%d%b%Y_%H%M%S") + "//"),
'Report' : ('report', 'report.html'),
I am trying to understand the working of "make" command (just started on this command). I have an ".sh" file which has a script to execute "make" command as shown below:
source /somepath/environment-setup-cortexa9hf-vfp-neon-poky-linux-gnueabi
make arch=arm toolchainPrefix=arm-poky-linux-gnueabi- xeno=off mode=Debug all
The directory where the script file is located has a file named "makefile". but there is nothing specified in the script file above regarding this "makefile". After executing the script file, all the script withing "makefile" is executed automatically. Can someone explain the working of "make xyz all" command in few words.
Thanks
As often with UNIX systems the command works to some degree by conventions. make (the GNU version of make at least) will search the working directory for files called GNUmakefile, makefile, and Makefile in that order or you can use the -f (or --file) option to give it a specific file.
I want a create a file with a specific extension(.done). I am using the command touch. Something Like:
touch `basename $UNZIPFILE`".done"
It's creating the file but in current directory. I want to create this file in a specific directory. Is there a option to provide the directory ?
I checked : http://ss64.com/bash/touch.html , but could not figure out.
I can think of one option is before this command I can do a cd requiredDIR
Is there any other way, I can specify the Directory on the same command, so that I dont have to change the Directory?
Simply prepend the directory variable to the file you are touching.
touch "$MYDIR/$(basename $UNZIPFILE).done"
If the directory doesn't exist, you need to create it.
mkdir -p "$MYDIR" && touch "$MYDIR/$(basename $UNZIPFILE).done"
(It's also better to use $(command) syntax instead of backticks for command substitution.)