The file names.txt consists of many names in the form of:
\"KELLEE\",\"JOSLYN\",\"JASON\",\"INGER\",\"INDIRA\",\"GLINDA\",\"GLENNIS\"
Does anyo
Jeremy's answer is basically correct and does exactly what you have asked for. But the format of your "names.txt" file is actually a well known and is called CSV (comma separated values). Luckily, Go comes with an encoding/csv package (which is part of the standard library) for decoding and encoding such formats easily. In addition to your + Jeremy's solution, this package will also give exact error messages if the format is invalid, supports multi-line records and does proper unquoting of quoted strings.
The basic usage looks like this:
package main
import (
"encoding/csv"
"fmt"
"io"
"os"
)
func main() {
file, err := os.Open("names.txt")
if err != nil {
fmt.Println("Error:", err)
return
}
defer file.Close()
reader := csv.NewReader(file)
for {
record, err := reader.Read()
if err == io.EOF {
break
} else if err != nil {
fmt.Println("Error:", err)
return
}
fmt.Println(record) // record has the type []string
}
}
There is also a ReadAll method that might make your program even shorter, assuming that the whole file fits into the memory.
Update: dystroy has just pointed out that your file has only one line anyway. The CSV reader works well for that too, but the following, less general solution should also be sufficient:
for {
if n, _ := fmt.Fscanf(file, "%q,", &name); n != 1 {
break
}
fmt.Println("name:", name)
}