在Go中将通用csv转换为xml

Question

I am trying to convert generic csv files into xml files. The csv file has a header line. The header values represent the element name, and the values in the respective column are the corresponding element values.

My approach so far:

// Read the csv file
file, err := os.Open(*i)
if err != nil {
  log.Fatalf("Error opening input file: %v
", err)
}
defer file.Close()

r := csv.NewReader(file)
r.Comma, _ = utf8.DecodeRuneInString(*s)
lines, err := r.ReadAll()

// header values
header := lines[0]
// Write the xml file
fileOut, err := os.Create(*o)
if err != nil {
  log.Fatalf("Error opening input file: %v
", err)
}
defer fileOut.Close()

for _, l := range lines {
  for i, elem := range l {
    // TODO: Something here is missing...
    xml.EscapeText(fileOut, []byte(header[i]))

    xml.EscapeText(fileOut, []byte(elem))

  }

Question: How do I make it the last part in xml write section work? Or is there a more efficient way of converting csv files into xml files?

Answer 1

The are two solutions: You can reflect the data into a struct,only when you can get the struct , fields and tags. then output XML by methods from xml package. But,if you don't know the struct as I face.there is a stupid but easy way,writing as string, as blow:

data := XML_HEADER
data = data + "<records>
"
for i := 2; i < len(records); i++ {
 data = data + "  <record "
  for j := 0; j < len(records[i]); j++  {
   data = data + head[j] + `="` + records[i][j] + `" `
                                        }
 data = data + "/>
"              }
 data = data + "</records>"
 ioutil.WriteFile(xmlFileName, []byte(data), 0644)