Capturing main process stdout into bash variable, keep printing subprocess output to console?

I'm developing a CLI. The CLI writes to stdout that I'd like to capture in a bash variable. That is, I'd like to be able to do this:

output=$(my_cli_command)

My CLI also runs a subprocess in that command. I'd like the output of the subprocess to be printed to the terminal when output=$(my_cli_command) is executed.

What do I have to do when running the subprocess to ensure that it (1) is printed to the console but (2) is not captured into the output variable? Or is this not possible?

I'm writing my CLI in Go. I've put a reproducible example here:

package main

import (
	"fmt"
	"os"
	"os/exec"
)

func main() {
	fmt.Println("I want this captured")

	// I don't want this captured, but I want "ls" printed to console
	command := exec.Command("ls")
	// With these lines, the print statement and the ls statement are printed to console.
	// All are captured in a bash variable.
	command.Stdout = os.Stdout
	command.Stderr = os.Stderr
	
	command.Run()

}

Copy that into a file main.go and run it like this:

output=$(go run main.go)
echo "$output"

Thank you.

This isn't a bash problem, it's a Go problem. If you don't want the output of ls to be captured, connect its stdout to your Go program's stderr rather than its stdout.

package main

import (
    "fmt"
    "os"
    "os/exec"
)

func main() {
    fmt.Println("I want this captured")

    command := exec.Command("ls")
    command.Stdout = os.Stderr // not os.Stdout
    command.Stderr = os.Stderr
    command.Run()
}


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# 答案2
**得分**: 1

清楚了，`ls` 命令的输出被捕获是因为它被发送到标准输出(stdout)。如果你运行的命令发送了任何内容到标准错误(stderr)，它将会在终端上显示而不会被捕获，就像你所希望的一样。例如，如果你将 Go 代码修改为以下形式：

    command := exec.Command("ls", "/no/such/file")

你将会得到以下结果：

     $ output=$(go run main.go)
     ls: /no/such/file: No such file or directory
     $ echo $output
     I want this captured

对于你的情况，解决方案是将子命令的标准输出连接到 `os.Stderr` 而不是 `os.Stdout`，就像 Charles Duffy 的回答中所示。

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<summary>英文:</summary>

To be clear, the `ls` output is captured _because it goes to stdout_. If you were to run a command that sent anything to stderr, it would be displayed on the terminal and not captured, just as you indicated you want. For example, if you change the Go code to do this:

    command := exec.Command(&quot;ls&quot; ,&quot;/no/such/file&quot;)

You get this result:

     $ output=$(go run main.go)
     ls: /no/such/file: No such file or directory
     $ echo $output
     I want this captured

The solution for your case is to connect the standard output of the child command to `os.Stderr` instead of `os.Stdout`, as in Charles Duffy&#39;s answer.

    



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