I think the best thing to do, if you're really concerned about the efficiency of concatenating all of these files, is to copy them all into the same bytes buffer.
buf := bytes.NewBuffer(nil)
for _, filename := range filenames {
f, _ := os.Open(filename) // Error handling elided for brevity.
io.Copy(buf, f) // Error handling elided for brevity.
f.Close()
}
s := string(buf.Bytes())
This opens each file, copies its contents into buf, then closes the file. Depending on your situation you may not actually need to convert it, the last line is just to show that buf.Bytes() has the data you're looking for.
If you just want the content as string
, then the simple solution is to use the ReadFile
function from the io/ioutil
package. This function returns a slice of bytes
which you can easily convert to a string
.
package main
import (
"fmt"
"io/ioutil"
)
func main() {
b, err := ioutil.ReadFile("file.txt") // just pass the file name
if err != nil {
fmt.Print(err)
}
fmt.Println(b) // print the content as 'bytes'
str := string(b) // convert content to a 'string'
fmt.Println(str) // print the content as a 'string'
}
I'm not with computer,so I write a draft. You might be clear of what I say.
func main(){
const dir = "/etc/"
filesInfo, e := ioutil.ReadDir(dir)
var fileNames = make([]string, 0, 10)
for i,v:=range filesInfo{
if !v.IsDir() {
fileNames = append(fileNames, v.Name())
}
}
var fileNumber = len(fileNames)
var contents = make([]string, fileNumber, 10)
wg := sync.WaitGroup{}
wg.Add(fileNumber)
for i,_:=range content {
go func(i int){
defer wg.Done()
buf,e := ioutil.Readfile(fmt.Printf("%s/%s", dir, fileName[i]))
defer file.Close()
content[i] = string(buf)
}(i)
}
wg.Wait()
}
This is how I did it:
package main
import (
"fmt"
"os"
"bytes"
"log"
)
func main() {
filerc, err := os.Open("filename")
if err != nil{
log.Fatal(err)
}
defer filerc.Close()
buf := new(bytes.Buffer)
buf.ReadFrom(filerc)
contents := buf.String()
fmt.Print(contents)
}
Source: Stackoverflow.com