


How to Properly Handle File Uploads in Multipart/Form-Data POST Requests with Go?
Troubleshooting Multipart/Form-Data POST in Go
In the context of sending a multipart form using Go's mime/multipart and http packages, you may encounter certain issues that require troubleshooting. Let's delve into a specific case related to creating a file field.
The Issue
You have the following code and are trying to send a multipart form with both text fields and a file:
func main() { // ... // Create a file field fw, err := w.CreateFormFile("file", "filename.dat") if err != nil { return err } // ... // Send the request resp, err := http.Post(url, w.FormDataContentType(), &buffer) if err != nil { return err } // ... }
However, you're struggling with how to obtain a Reader to read the file into the file field writer fw.
The Solution
To solve this issue, you need to open the file using os.Open and pass the returned *os.File as the io.Writer to the CreateFormFile function. Here's the corrected code:
import ( "bytes" "io" "mime/multipart" "net/http" "os" ) func main() { // ... // Open the file fd, err := os.Open("filename.dat") if err != nil { return err } // Create a file field fw, err := w.CreateFormFile("file", "filename.dat") if err != nil { return err } // Copy the file contents into the file field writer _, err = io.Copy(fw, fd) if err != nil { return err } // ... // Send the request resp, err := http.Post(url, w.FormDataContentType(), &buffer) if err != nil { return err } // ... }
With this modification, the file's contents will be correctly written into the multipart form, allowing you to send both text fields and a file successfully.
The above is the detailed content of How to Properly Handle File Uploads in Multipart/Form-Data POST Requests with Go?. For more information, please follow other related articles on the PHP Chinese website!

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