I am trying to change the directory in the command line from a gen_server using
os:cmd("cd d:\temp").
but nothing happense, the return is just an empty list and I remain in the same directory.
any ideas?
Try using file:set_cwd(Dir) to change your current dir.
cmd() runs a sub-shell, which you're telling to change directory, then the sub-shell exits, having changed nothing about its parent process's environment.
You want to use cd() instead, if you're in the shell, or file:set_cwd() at runtime within an Erlang program.
Another option, if you want to run another program and have its working directory be different from the one Erlang is using is to pass the {cd, Dir} tuple to open_port().
Related
I am currenly working on an erlang project and stuck in reading the file. I want to read a text file which is in the /src folder where all the erlang and a text file are in the same structure. Then too, I am not being able to read the file despite of specifying file paths. Any help would be appreciated.
start() ->
{ok,DataList} = file:consult("Calls.txt"),
io:format("** Calls to be made **"),
io:fwrite("~w~n",[DataList]).
The data file stores contents like : {john, [jill,joe,bob]}.
Try add folder name to the path or try set full patch to the file:
1> {ok,DataList} = file:consult("src/Calls.txt").
Notes: the error {error,enoent} mean that the file does not exist or you don't have a rights to read/write current file, for this case need set 777 rights or similar.
If you need to use src/call.txt, then this simply means that your IDE (or you) has created a src folder in which the calls.txt file has been placed. At the same time, the IDE is using a path that only includes the top level folder (i.e., the root folder for the IDE project). So src/call.txt must be used in that case. This isn’t a problem with Erlang, or even the IDE. It’s just the way your project is set up.
You can do either of two things. Move the calls.txt file up one level in the IDE file manager, so that it can be referenced as calls.txt, not src/call.txt. You can also just change the path to “calls.txt” before you run it from the command line.
enoent means "Error: No Entry/Entity". It means the file couldn't be found. When I try your code, it works correctly and outputs
[{john,[jill,joe,bob]}]
How can I retrieve the path to the go binary in a container that does not have which programatically?
One option would be to execute which go as follows:
bytes, err := exec.Command("which", "go").Output()
However, I would like to not depend on which being available. Does go provide any in-built mechanism to retrieve this and if not, what is the alternative apart from having the user pass in the path themselves?
From the man page of which:
Which takes one or more arguments. For each of its arguments it prints to stdout the full path of the executables that would have been executed when this argument had been entered at the shell prompt. It does this by searching for an executable or script in the directories listed in the environment variable PATH using the same algorithm as bash(1).
Go's os/exec.LookPath function is very close to this functionality:
LookPath searches for an executable named file in the directories named by the PATH environment variable. If file contains a slash, it is tried directly and the PATH is not consulted. The result may be an absolute path or a path relative to the current directory.
Use path/filepath.Abs if you need a guaranteed absolute path.
I don't expect this to be the best answer, but it is one that I just found. I was hoping for more of a go-specific one, but in the meantime type in linux is a default built-in one available in bash and sh (alpine).
You can test this yourself by running type type which yields:
type is a shell builtin
The usage in go would look like this:
b, err := exec.Command("type", "go").Output()
if err != nil {
/* 'type' is not available on the O/S */
}
goPath := strings.TrimPrefix(strings.TrimSuffix(string(b), "\n"), "go is ")
The reason the Trim functions are required is because the output would look like this:
go is /usr/local/go/bin/go\n
This isn't the nicest way of going about it, but it works.
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.
What is the difference between:
include("./somepath/class.php");
and
include("somepath/class.php");
There shouldn't be any difference, directory wise, since the former assumes relative directory structure. The only potential pitfall could be if "somepath" were actually a command to be run - some users expect to type a command for a local script file and assume it should run, when you actually have to run "./somepath" to invoke it. This, of course, only pertains to commands not on your $PATH.
I don't see any difference. "." means current directory.
. refers to the current working directory.
Sometimes you want to specify explicitly that what you want is in the current working directory, and that it is not from something in your path variable for example.
For example in PHP "somepath/somefile" is appended after paths specified in include_dir (for example: /home/var/; /home/bin/ ....) directive.
The second variant is more specific it says: search in current directory!
I make a call just like this:
value = ./simulated_annealing
Which is a C Object file, but Rails tells me it cannot find that file. I
put it in the same dir that the rest of the models files (since it's
called by one of those models), but I guess it should be in any other
place.
I've tried that outside Ruby and it works great.
What do I have to do?
The thing is, when you say:
./simulated_annealing
you're explicitly saying: run file named simulated_annealing which is found in the current directory. That's what the ./ means. If the file's located elsewhere you need to provide the path to it, or add that path to the environment variable $PATH. So, you should replace that line with:
/path/to/simulated_annealing
where /path/to represents the actual path.
The best option is to use an absolute path for running the program. For ex.,
you can create a directory "bin" under your rails application top level
directory. Place your program under "bin" directory. Then you can
execute the program something like:
cmd = "#{RAILS_ROOT}/bin/cbin arg1 arg2"
value = `#{cmd}`