Golang file upload to s3 using OS Open - file-upload

Question

I am trying to upload an Image to my s3 account using Golang and the amazon s3 api . I can get the imagine uploaded if I hard code the direct path such as
file, err := os.Open("/Users/JohnSmith/Documents/pictures/cars.jpg")
defer file.Close()
if err != nil {
fmt.Printf("err opening file: %s", err)
}
if I hard code the file path like that then the picture will be uploaded to my s3 account . However that approach is not good as I can't obviously hard code the direct image path to every image that I want to upload . My question is how can I upload images without having to Hardcode the path . This will be apart of an API where users will upload images so I clearly can not have a hard coded path . This is my code first the HTML
<form method="post" enctype="multipart/form-data" action="profile_image">
<h2>Image Upload</h2>
<p><input type="file" name="file" id="file"/> </p>
<p> <input type="submit" value="Upload Image"></p>
</form>
then this is my HTTP Post function method
func UploadProfile(w http.ResponseWriter, r *http.Request) {
r.ParseForm()
var resultt string
resultt = "Hi"
sess, _ := session.NewSession(&aws.Config{
Region: aws.String("us-west-2"),
Credentials: credentials.NewStaticCredentials(aws_access_key_id,aws_secret_access_key, ""),
})
svc := s3.New(sess)
file, err := os.Open("Users/JohnSmith/Documents/pictures/cars.jpg")
defer file.Close()
if err != nil {
fmt.Printf("err opening file: %s", err)
}
fileInfo, _ := file.Stat()
size := fileInfo.Size()
buffer := make([]byte, size) // read file content to buffer
file.Read(buffer)
fileBytes := bytes.NewReader(buffer)
fileType := http.DetectContentType(buffer)
path := file.Name()
params := &s3.PutObjectInput{
Bucket: aws.String("my-bucket"),
Key: aws.String(path),
Body: fileBytes,
ContentLength: aws.Int64(size),
ContentType: aws.String(fileType),
}
resp, err := svc.PutObject(params)
if err != nil {
fmt.Printf("bad response: %s", err)
}
fmt.Printf("response %s", awsutil.StringValue(resp))
}
That is my full code above however when I try to do something such as
file, err := os.Open("file")
defer file.Close()
if err != nil {
fmt.Printf("err opening file: %s", err)
}
I get the following error
http: panic serving [::1]:55454: runtime error: invalid memory address or nil pointer dereference
goroutine 7 [running]:
err opening file: open file: no such file or directorynet/http.(*conn).serve.func1(0xc420076e80)
I can't use absolute path (filepath.Abs()) because some of the files will be outside of the GoPath and as stated other users will be uploading. Is there anyway that I can get a relative path ..

Answer 1

After POST to your API, images are temporarily saved in a OS's temp directory (different for different OS's) by default. To get this directory you can use, for example:
func GetTempLoc(filename string) string {
return strings.TrimRight(os.TempDir(), "/") + "/" + filename
}
Where:
filename is a header.Filename, i.e. file name received in your POST request. In Gin-Gonic framework you get it in your request handler as:
file, header, err := c.Request.FormFile("file")
if err != nil {
return out, err
}
defer file.Close()
Example: https://github.com/gin-gonic/gin#another-example-upload-file.
I'm sure in your framework there will be an analogue.
os.TempDir() is a function go give you a temp folder (details: https://golang.org/pkg/os/#TempDir).
TrimRight is used to ensure result of os.TempDir is consistent on different OSs
And then you use it as
file, err := os.Open(GetTempLoc(fileName))
...