shell腳本中如何判斷umount的成功/失敗

wooin

shell腳本中應(yīng)該如何表達這個意思呢？

if [ umount /mnt/disk ]; then
      echo "umount done!"
fi

謝謝！

ly5066113

回復(fù) 1# wooin

1. umount /mnt/disk && echo "umount done!"

復(fù)制代碼

或者：

1. umount /mnt/disk
   
2. if [ $? -eq 0 ]; then
   
3.       echo "umount done!"
   
4. fi

復(fù)制代碼

waker

if 　umount /mnt/disk
then
      echo "umount done!"
fi


umount /mnt/disk&&echo "umount done!"

haokoo

回復(fù) 3# waker


    umount 執(zhí)行成功不是應(yīng)該返回0嘛？  那if的判斷應(yīng)該不會執(zhí)行吧？


測試了一下，判斷通過...為毛

waker

回復(fù) 4# haokoo


    PAGER='less -iRsp "exit status"' man bash

liyunxiang12

回復(fù) 4# haokoo
shell中 0 也是true


   

haokoo

回復(fù) 5# waker

1. if list; then list; [ elif list; then list; ] ... [ else list; ] fi
   
2.               The if list is executed.  If its exit status is zero, the then list  is  executed.   Other-
   
3.               wise, each elif list is executed in turn, and if its exit status is zero, the corresponding
   
4.               then list is executed and the command completes.  Otherwise, the else list is executed,  if
   
5.               present.   The  exit  status is the exit status of the last command executed, or zero if no
   
6.               condition tested true.

復(fù)制代碼

看到了，謝謝提醒。