How to use gRPC to implement file upload in Golang?

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:

// 文件上传服务接口
syntax = "proto3";

package file_upload;

service FileUploadService {
  rpc Upload(stream FileChunk) returns (FileUploadResponse);
}

// 文件块信息
message FileChunk {
  bytes data = 1;
}

// 文件上传响应
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() {
    // 建立与服务的连接
    conn, err := grpc.Dial("127.0.0.1:50051", grpc.WithInsecure())
    if err != nil {
        log.Fatalf("连接服务失败: %v", err)
    }
    defer conn.Close()

    // 创建文件上传客户端
    client := file_upload.NewFileUploadServiceClient(conn)

    // 打开要上传的文件
    file, err := os.Open("example.txt")
    if err != nil {
        log.Fatalf("打开文件失败: %v", err)
    }
    defer file.Close()

    // 将文件分成块进行流式传输
    stream, err := client.Upload(context.Background())
    if err != nil {
        log.Fatalf("创建流式传输失败: %v", err)
    }

    // 循环读取文件并发送块
    buf := make([]byte, 1024)
    for {
        n, err := file.Read(buf)
        if err == io.EOF {
            break
        }
        if err != nil {
            log.Fatalf("读取文件失败: %v", err)
        }

        // 发送块
        if err := stream.Send(&file_upload.FileChunk{Data: buf[:n]}); err != nil {
            log.Fatalf("发送块失败: %v", err)
        }
    }

    stream.CloseSend()

    // 接收上传响应
    resp, err := stream.Recv()
    if err != nil {
        log.Fatalf("接收上传响应失败: %v", err)
    }

    log.Printf("上传成功: %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() {
    // 创建文件上传服务
    server := &FileUploadService{}

    // 创建 gRPC 服务器并注册服务
    s := grpc.NewServer()
    file_upload.RegisterFileUploadServiceServer(s, server)

    lis, err := net.Listen("tcp", "127.0.0.1:50051")
    if err != nil {
        log.Fatalf("监听端口失败: %v", err)
    }

    // 启动 gRPC 服务器
    if err := s.Serve(lis); err != nil {
        log.Fatalf("启动 gRPC 服务器失败: %v", err)
    }
}

// FileUploadService 服务的具体实现
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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