问题描述
根据golang文档，当您使用exec.Command()时，go不会调用系统的外壳程序.
As per the golang documentation, go does not make a call to the system's shell when you are using exec.Command().
来自"os/exec"软件包中的golang.org文档:
From the golang.org documentation on the "os/exec" package:
这带来了一个问题.由于这种设计选择，您在执行命令时不能使用管道.因此，以下代码无法按预期执行.
This presents a problem. Because of this design choice you cannot use piping when executing a command. Therefore the following code does not execute as desired.
package main
import (
"fmt"
"os/exec"
)
func main() {
exec.Command("echo", "Hello", ">>", "~/thing").Run()
cmdOut, _ := exec.Command("cat", "~/thing").Output()
fmt.Println(cmdOut)
}
与其打印出应该包含单词"Hello"的文件内容，不如输出空白的换行符.我已经尝试过像这样直接调用bash:
Instead of printing out the contents of a file that should contain the word 'Hello,' it instead prints out a blank newline. I have tried directly invoking bash like this:
package main
import (
"fmt"
"os/exec"
)
func main() {
exec.Command("bash", "-c", "echo", "Hello", ">>", "~/thing").Run()
cmdOut, _ := exec.Command("cat", "~/thing").Output()
fmt.Println(cmdOut)
}
但是，这将产生与原始代码相同的结果.使用golang时如何直接调用系统外壳?
This, however, produces the same result as the original code. How can I directly invoke the system shell when using golang?
推荐答案
第二个参数应该是一个字符串.在shell命令中，您也需要将其作为一个字符串传递. bash也解释~.您可以放心地假设sh存在. Bash shell不是必须的.
The second argument should be one string. In shell command you need to pass it as one string too. Also ~ is interpreted by bash. You can safely assume that sh exists. Bash shell is not a must.
package main
import (
"fmt"
"os/exec"
)
func main() {
exec.Command("sh", "-c", "echo Hello >> ~/thing").Run()
cmdOut, _ := exec.Command("sh", "-c", "cat ~/thing").Output()
fmt.Println(cmdOut)
}
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