How to Convert a CSV File to XML in Go
The CSV format is one of the most predominant players when it comes to data organization, but it can raise some problems with readability, especially if you don’t have knowledge of the column headers. If you need a more user/web-friendly format, look no further than an XML file. The following API will allow you to easily transform your CSV document into XML with no additional stress; the parameters only require the input file with an option for specifying the use of column names.
We are going to perform the document conversion by instancing the API and calling the function:
package mainimport (
"fmt"
"bytes"
"mime/multipart"
"os"
"path/filepath"
"io"
"net/http"
"io/ioutil"
)func main() {url := "https://api.cloudmersive.com/convert/csv/to/xml"
method := "POST"payload := &bytes.Buffer{}
writer := multipart.NewWriter(payload)
file, errFile1 := os.Open("/path/to/file")
defer file.Close()
part1,
errFile1 := writer.CreateFormFile("inputFile",filepath.Base("/path/to/file"))
_, errFile1 = io.Copy(part1, file)
if errFile1 != nil {
fmt.Println(errFile1)
return
}
err := writer.Close()
if err != nil {
fmt.Println(err)
return
}client := &http.Client {
}
req, err := http.NewRequest(method, url, payload)if err != nil {
fmt.Println(err)
return
}
req.Header.Add("columnNamesFromFirstRow", "<boolean>")
req.Header.Add("Content-Type", "multipart/form-data")
req.Header.Add("Apikey", "YOUR-API-KEY-HERE")req.Header.Set("Content-Type", writer.FormDataContentType())
res, err := client.Do(req)
if err != nil {
fmt.Println(err)
return
}
defer res.Body.Close()body, err := ioutil.ReadAll(res.Body)
if err != nil {
fmt.Println(err)
return
}
fmt.Println(string(body))
}
After the operation is complete, you will have your newly created XML file!