How to use gRPC to implement file upload in Golang?
Jun 03, 2024 pm 04:54 PM
How to use gRPC to upload files? Create supporting service definitions, including request and response messages. On the client, the file to be uploaded is opened and split into chunks, then streamed to the server via a gRPC stream. On the server side, file chunks are received and stored into a file. The server sends a response after the file upload is completed to indicate whether the upload was successful.
How to use gRPC to implement file upload in Golang
gRPC (Google Remote Procedure Call) is a general-purpose, high-performance An open source remote procedure call framework that can run in any environment. It provides an efficient way to transfer data between different machines and is ideal for situations where large files need to be transferred in a distributed system.
Supporting Service
Before you begin, you need to create a supporting service to handle file uploads. The following is a sample service definition:
// 文件上傳服務(wù)接口
syntax = "proto3";
package file_upload;
service FileUploadService {
rpc Upload(stream FileChunk) returns (FileUploadResponse);
}
// 文件塊信息
message FileChunk {
bytes data = 1;
}
// 文件上傳響應(yīng)
message FileUploadResponse {
bool success = 1;
}gRPC client implementation
The file upload process using gRPC client implementation is as follows:
package main
import (
"context"
"io"
"log"
file_upload "github.com/example/file-upload"
"google.golang.org/grpc"
)
func main() {
// 建立與服務(wù)的連接
conn, err := grpc.Dial("127.0.0.1:50051", grpc.WithInsecure())
if err != nil {
log.Fatalf("連接服務(wù)失敗: %v", err)
}
defer conn.Close()
// 創(chuàng)建文件上傳客戶端
client := file_upload.NewFileUploadServiceClient(conn)
// 打開要上傳的文件
file, err := os.Open("example.txt")
if err != nil {
log.Fatalf("打開文件失敗: %v", err)
}
defer file.Close()
// 將文件分成塊進(jìn)行流式傳輸
stream, err := client.Upload(context.Background())
if err != nil {
log.Fatalf("創(chuàng)建流式傳輸失敗: %v", err)
}
// 循環(huán)讀取文件并發(fā)送塊
buf := make([]byte, 1024)
for {
n, err := file.Read(buf)
if err == io.EOF {
break
}
if err != nil {
log.Fatalf("讀取文件失敗: %v", err)
}
// 發(fā)送塊
if err := stream.Send(&file_upload.FileChunk{Data: buf[:n]}); err != nil {
log.Fatalf("發(fā)送塊失敗: %v", err)
}
}
stream.CloseSend()
// 接收上傳響應(yīng)
resp, err := stream.Recv()
if err != nil {
log.Fatalf("接收上傳響應(yīng)失敗: %v", err)
}
log.Printf("上傳成功: %v", resp.Success)
}Supporting service implementation
The server implements the logic of receiving and processing uploaded files:
package main
import (
"context"
"io"
"log"
file_upload "github.com/example/file-upload"
)
func main() {
// 創(chuàng)建文件上傳服務(wù)
server := &FileUploadService{}
// 創(chuàng)建 gRPC 服務(wù)器并注冊(cè)服務(wù)
s := grpc.NewServer()
file_upload.RegisterFileUploadServiceServer(s, server)
lis, err := net.Listen("tcp", "127.0.0.1:50051")
if err != nil {
log.Fatalf("監(jiān)聽端口失敗: %v", err)
}
// 啟動(dòng) gRPC 服務(wù)器
if err := s.Serve(lis); err != nil {
log.Fatalf("啟動(dòng) gRPC 服務(wù)器失敗: %v", err)
}
}
// FileUploadService 服務(wù)的具體實(shí)現(xiàn)
type FileUploadService struct{}
func (s *FileUploadService) Upload(stream file_upload.FileUploadService_UploadServer) error {
file, err := os.Create("received.txt")
if err != nil {
return err
}
for {
chunk, err := stream.Recv()
if err == io.EOF {
break
}
if err != nil {
return err
}
if _, err := file.Write(chunk.GetData()); err != nil {
return err
}
}
if err := stream.SendAndClose(&file_upload.FileUploadResponse{Success: true}); err != nil {
return err
}
return nil
}The above code shows how to use gRPC to implement file upload in Golang. The client breaks the file into chunks and streams them to the server, which is responsible for receiving and processing the uploaded file.
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