golang os/exec调用shell脚本卡住问题排
2016-03-15 | pyinx | golang

代码如下

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package main

import (
        "fmt"
        "os/exec"
)

func main() {
        cmd := exec.Command("/bin/sh", "/tmp/1.sh")    
        out, err := cmd.Output()                       

        if err != nil {                                
                fmt.Println(err)                       
        } else {                                       
                fmt.Println(string(out))               
        }                                              
}
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# cat /tmp/1.sh
date
#./t  （不会卡住）
Tue Mar 15 17:59:29 CST 2016
# vim /tmp/1.sh
# cat /tmp/1.sh
/etc/init.d/salt-minion restart
# strace ./t （这里一直卡住不退出）
...
read(4, "Stopping salt-minion daemon: ", 512) = 29
futex(0x58bac0, FUTEX_WAKE, 1)          = 1
read(4, "[", 1507)                      = 1
futex(0x58bac0, FUTEX_WAKE, 1)          = 1
read(4, "  OK  ]\r", 1506)              = 8
read(4, "\n", 1498)                     = 1
futex(0x58bac0, FUTEX_WAKE, 1)          = 1
read(4, "Starting salt-minion daemon: ", 1497) = 29
read(4, "[", 1468)                      = 1
futex(0x58bac0, FUTEX_WAKE, 1)          = 1
futex(0x58ba40, FUTEX_WAKE, 1)          = 1
read(4, "  OK  ]\r\n", 1467)            = 9
read(4, 0xc20807a04e, 1458)             = ? ERESTARTSYS (To be restarted)
--- SIGWINCH (Window changed) @ 0 (0) ---
rt_sigreturn(0x1c)                      = 0
read(4,

执行/etc/init.d/salt-minion restart后，脚本一直不退出。
困扰了我一下午，一直找不到原因，后来找组内的go浪大神帮忙看了下，问题终于解决了。

原因

执行/etc/init.d/salt-minion restart后，会启动一个salt-minion的子进程.cmd.Output执行后会去调一个Wait的方法，Wait方法会等待所有进程都执行完后才会退出。这样就形成了死循环，导致一直不退出。

解决

直接调用exec的Start方法，绕过Wait方法。

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package main

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

func main() {
        var b bytes.Buffer
        command := exec.Command("/bin/sh", "/tmp/1.sh")
        command.Stdout = &b
        command.Stderr = &b
        err := command.Start()
        if err != nil {
                fmt.Println(b.String(), err)
        }
        _, err = command.Process.Wait() //等待主进程执行完成
        fmt.Println(b.String(), err)
}