How to get the directory of the currently running file?

This should do it:

import (
    "fmt"
    "log"
    "os"
    "path/filepath"
)

func main() {
    dir, err := filepath.Abs(filepath.Dir(os.Args[0]))
    if err != nil {
            log.Fatal(err)
    }
    fmt.Println(dir)
}

Use package osext

It's providing function ExecutableFolder() that returns an absolute path to folder where the currently running program executable reside (useful for cron jobs). It's cross platform.

Online documentation

package main

import (
    "github.com/kardianos/osext"
    "fmt"
    "log"
)

func main() {
    folderPath, err := osext.ExecutableFolder()
    if err != nil {
        log.Fatal(err)
    }
    fmt.Println(folderPath)
}

EDIT: As of Go 1.8 (Released February 2017) the recommended way of doing this is with os.Executable:

func Executable() (string, error)

Executable returns the path name for the executable that started the current process. There is no guarantee that the path is still pointing to the correct executable. If a symlink was used to start the process, depending on the operating system, the result might be the symlink or the path it pointed to. If a stable result is needed, path/filepath.EvalSymlinks might help.

To get just the directory of the executable you can use path/filepath.Dir.

Example:

package main

import (
    "fmt"
    "os"
    "path/filepath"
)

func main() {
    ex, err := os.Executable()
    if err != nil {
        panic(err)
    }
    exPath := filepath.Dir(ex)
    fmt.Println(exPath)
}

OLD ANSWER:

You should be able to use os.Getwd

func Getwd() (pwd string, err error)

Getwd returns a rooted path name corresponding to the current directory. If the current directory can be reached via multiple paths (due to symbolic links), Getwd may return any one of them.

For example:

package main

import (
    "fmt"
    "os"
)

func main() {
    pwd, err := os.Getwd()
    if err != nil {
        fmt.Println(err)
        os.Exit(1)
    }
    fmt.Println(pwd)
}

I came from Node.js to Go. The Node.js equivalent to __dirname in Go is:

_, filename, _, ok := runtime.Caller(0)
if !ok {
    return errors.New("unable to get the current filename")
}
dirname := filepath.Dir(filename)

Some other mentions in this thread and why they're wrong:

  • os.Executable() will give you the filepath of the currently running executable. This is equivalent to process.argv[0] in Node. This is not true if you want to take the __dirname of a sub-package.
  • os.Getwd() will give you the current working directory. This is the equivalent to process.cwd() in Node. This will be wrong when you run your program from another directory.

Lastly, I'd recommend against pulling in a third-party package for this use case. Here's a package you can use:

package current

// Filename is the __filename equivalent
func Filename() (string, error) {
    _, filename, _, ok := runtime.Caller(1)
    if !ok {
        return "", errors.New("unable to get the current filename")
    }
    return filename, nil
}


// Dirname is the __dirname equivalent
func Dirname() (string, error) {
    filename, err := Filename()
    if err != nil {
        return "", err
    }
    return filepath.Dir(filename), nil
}

Note that I've adjusted runtime.Caller(1) to 1 because we want to get the directory of the package that called current.Dirname(), not the directory containing the current package.

Tags:

Go

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