Say I have a file on my desktop that doesn't have extension, but contains some code (say Batch or VBS).
Can I run this code without adding an extension to the file?
I was personally thinking about using powershell to pass an extension to wscript.exe, but this turned out to be somewhat impossible.
For VBS\JS scripts you need to specify engine with //E switch.
Examples:
If file xxx contains VbScript code: wscript //E:VBScript c:\xxx
If file xxx contains JavaScript code: wscript //E:JScript c:\xxx
For batch files, it's not so easy. There is some workarounds, but they're somewhat limited: How to run batch script with out using *.bat extension
Related
I want to create some "build traceability" functionality, and include the actual bazel command that was run to produce one of my build artifacts. So if the user did this:
bazel run //foo/bar:baz --config=blah
I want to actually get the string "bazel run //foo/bar:baz --config=blah" and write that to a file during the build. Is this possible?
Stamping is the "correct" way to get information like that into a Bazel build. Note the implications around caching though. You could also just write a wrapper script that puts the command line into a file or environment variable. Details of each approach below.
I can think of three ways to get the information you want via stamping, each with differing tradeoffs.
First way: Hijack --embed_label. This shows up in the BUILD_EMBED_LABEL stamping key. You'd add a line like build:blah --embed_label blah in your .bazelrc. This is easy, but that label is often used for things like release_50, which you might want to preserve.
Second way: hijack the hostname or username. These show up in the BUILD_HOST and BUILD_USER stamping keys. On Linux, you can write a shell script at tools/bazel which will automatically be used to wrap bazel invocations. In that shell script, you can use unshare --uts --map-root-user, which will work if the machine is set up to enable bazel's sandboxing. Inside that new namespace, you can easily change the hostname and then exec the real bazel binary, like the default /usr/bin/bazel shell script does. That shell script has full access to the command line, so it can encode any information you want.
Third way: put it into an environment variable and have a custom --workspace_status_command that extracts it into a stamping key. Add a line like build:blah --action_env=MY_BUILD_STYLE=blah to your .bazelrc, and then do echo STABLE_MY_BUILD_STYLE ${MY_BUILD_STYLE} in your workspace status script. If you want the full command line, you could have a tools/bazel wrapper script put that into an environment variable, and then use build --action_env=MY_BUILD_STYLE to preserve the value and pass it to all the actions.
Once you pick a stamping key to use, src/test/shell/integration/stamping_test.sh in the Bazel source tree is a good example of writing stamp information to a file. Something like this:
genrule(
name = "stamped",
outs = ["stamped.txt"],
cmd = "grep BUILD_EMBED_LABEL bazel-out/volatile-status.txt | cut -d ' ' -f 2 >\$#",
stamp = True,
)
If you want to do it without stamping, just write the information to a file in the source tree in a tools/bazel wrapper. You'd want to put that file in your .gitignore, of course. echo "$#" > cli_args is all it takes to dump them to a file, and then you can use that as a source file like normal in your build. This approach is simplest, but interacts the most poorly with Bazel's caching, because everything that depends on that file will be rebuilt every time with no way to control it.
I'm trying to list the files of a directory without using the systems shell like it is done by using io.popen or os.execute.
So, the following does not work:
local content = io.popen("ls -la", "r")
Background: the interpreter is stripped down and does not support the above mentioned functions in the standard libraries.
Is there any other small and simple way to perform a directory listing in Lua?
I would like to add this to my .bazelrc, but the $(whoami) doesn't expand like if it was in a shell.
startup --output_user_root=/tmp/bazel/out/$(whoami)
It produces the literal result:
/tmp/bazel/out/$(whoami)/faedb999bdce730c9c495251de1ca1a4/execroot/__main__/bazel-out/
Is there any way to do what I want: adding a name/hash to the option in the .bashrc file?
Edit: what I really want is to set the outputRoot to /tmp/bazel/out without using an environment variable and to let bazel create it's user and workspace hash directories there.
You can run Bazel from a wrapper script. In fact, that's exactly what the bazel binary is (at least on Linux): it's a wrapper script that calls bazel-real. You can edit this wrapper script if you like, or rename it to bazel.sh and write your own wrapper.
/usr/bin/bazel is a script which looks for //tools/bazel, and if it exists, calls it. Otherwise, it calls bazel-real. This lets you check Bazel into your repo, or otherwise modify how it gets called. We use that to download a specific version of bazel, extract it, and then call it.
I would recommend creating //tools/bazel, and having that do your modification. It can then either call a versioned version of bazel, or call bazel-real. That keeps your modifications local to your repo rather than global.
I have a spss syntax file that I need to run on multiple files each in a different directory with the same name as the file, and I am trying to too do this automatically. So far I have tried doing it with syntax code and am trying to avoid doing python is spss, but all I have been able to get is the code bellow which does not work.
VECTOR v = key.
LOOP #i = 1 to 41.
GET
FILE=CONCAT('C:\Users\myDir\otherDir\anotherDir\output\',v(#i),'\',v(#i),'.sav').
DATASET NAME Data#i WINDOW=FRONT.
*Do stuff to the opened file
END LOOP.
EXE.
key is the only column in a file that contains all the names of the files.
I am having trouble debugging since I don't know how to print to the screen if it is possible. So my question is: is there a way to get the code above to work, or another option that accomplishes the same thing?
You can't use an expression like that on a GET command. There are two choices. Use the macro language to put this together (see DEFINE in the Command Syntax Reference via the Help menu) or use the SPSSINC PROCESS FILES extension command or your own Python code to select the files with a wildcard.
The extension command or a Python program require the free Python Essentials available from the SPSS Community website or available with your Statistics version.
Lets say, I setup my svn host, like: http://www.example.com/svn
Then I create my project at a folder like: /home/me/workspace/my_app
I want to Setup my svn client and ignore files like tmp/* log/* db/schema.rb and such, so what are the commands on console for that? If possible can you write what the full sequence of commands to set this up?
obs: I have found no GUI that suit my needs on Ubuntu kdesvn / rapidsvn :(
Using TortoiseSVN, you could simply use the ignore menu item, but I guess you're not using a GUI.
You can ignore single files by executing this in the containing directory:
svn propset svn:ignore filename .
Replace filename by the filename you want to ignore. This can also be a wildcard expression like *.suo. It can also be a directory like tmp.
you can define the prop svn:ignore
so in your directory where you want ignore file made :
svn propedit svn:ignore .
An editor is open an add your information into. Close it and your file are ignore