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Opening a file by name

  • File access is a very common operation used to store or read the data. This recipe illustrates how to open a file by its name and path, using the standard library.

Create the directory temp and create the file file.txt in it.

Edit the file.txt file and write This file content into the file.

Create the openfile.go file with the following content:

        package main

        import (
          "fmt"
          "io"
          "io/ioutil"
          "os"
        )

        func main() {

          f, err := os.Open("temp/file.txt")
          if err != nil {
            panic(err)
          }

          c, err := ioutil.ReadAll(f)
          if err != nil {
            panic(err)
          }

          fmt.Printf("### File content ###\n%s\n", string(c))
          f.Close()

          f, err = os.OpenFile("temp/test.txt", os.O_CREATE|os.O_RDWR,
                               os.ModePerm)
          if err != nil {
            panic(err)
          }
          io.WriteString(f, "Test string")
          f.Close()

        }


output:

sangam:golang-daily sangam$ go run openfile.go
### File content ###
This file content

sangam:golang-daily sangam$ 

See the output there should also be a new file, test.txt, in the temp folder:

How it works…

  • The os package offers a simple way of opening the file. The function Open opens the file by the path, just in read-only mode. Another function, OpenFile, is the more powerful one and consumes the path to the file, flags, and permissions.

  • The flag constants are defined in the os package and you can combine them with use of the binary OR operator . The permissions are set by the os package constants (for example, os.ModePerm ) or by the number notation such as 0777 (permissions: -rwxrwxrwx).