rmo*_*njo 35 http flush stream go
我正在Go中编写一个简单的Web应用程序,我希望将我的响应流式传输到客户端(即,在请求完全处理后,不会缓冲并以块的形式发送):
func handle(res http.ResponseWriter, req *http.Request) {
fmt.Fprintf(res, "sending first line of data")
sleep(10) //not real code
fmt.Fprintf(res, "sending second line of data")
}
Run Code Online (Sandbox Code Playgroud)
从客户端的角度来看,这两行将同时发送.任何建议值得赞赏:)
在我亲自制作的每次写入后都可以刷新,但在我的用例中,这还不够:
cmd := exec.Command("a long command that outputs lots of lines")
cmd.Stdout = res //where res is a http.ResponseWritter
cmd.Stderr = res
err := cmd.Run()
Run Code Online (Sandbox Code Playgroud)
我希望我的输出cmd也被刷新.无论如何要"autoflush"ResponseWritter?
我在golang的邮件列表上找到了帮助.有两种方法可以实现这一点:使用允许接管HTTP的底层TCP连接的劫持程序,或者在写入和刷新的go例程中管道命令的stdout和stderr:
pipeReader, pipeWriter := io.Pipe()
cmd.Stdout = pipeWriter
cmd.Stderr = pipeWriter
go writeCmdOutput(res, pipeReader)
err := cmd.Run()
pipeWriter.Close()
//---------------------
func writeCmdOutput(res http.ResponseWriter, pipeReader *io.PipeReader) {
buffer := make([]byte, BUF_LEN)
for {
n, err := pipeReader.Read(buffer)
if err != nil {
pipeReader.Close()
break
}
data := buffer[0:n]
res.Write(data)
if f, ok := res.(http.Flusher); ok {
f.Flush()
}
//reset buffer
for i := 0; i < n; i++ {
buffer[i] = 0
}
}
}
Run Code Online (Sandbox Code Playgroud)
Den*_*ret 28
正如文档中暗示的那样,有些人ResponseWriter可能会实现该Flusher接口.
这意味着您可以执行以下操作:
func handle(res http.ResponseWriter, req *http.Request) {
fmt.Fprintf(res, "sending first line of data")
if f, ok := res.(http.Flusher); ok {
f.Flush()
} else {
log.Println("Damn, no flush");
}
sleep(10) //not real code
fmt.Fprintf(res, "sending second line of data")
}
Run Code Online (Sandbox Code Playgroud)
请注意,在网络或客户端的许多其他位置可能会发生缓冲.