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linuxbashsudowatchdog

syntax error near unexpected token `do' when run with sudo

From here: http://www.sat.dundee.ac.uk/psc/watchdog/watchdog-testing.html

for n in $(seq 1 60); do echo $n; sleep 1; sync; done

I get:

:~$ sudo for n in $(seq 1 60); do echo $n; sleep 1; sync; done  
bash: syntax error near unexpected token `do'
Solution

  • The shell parses the command line and because for looks like an argument to sudo, you basically get a do without a for.

    To fix it, run the loop in a subshell, either as a separate script, or like this;

    sudo sh -c 'for n in $(seq 1 60); do echo "$n"; sleep 1; sync; done'

    Better yet, avoid running anything unnecessary as a privileged user:

    for n in $(seq 1 60); do echo "$n"; sleep 1; sudo sync; done

    The first sudo will require a password, but subsequent iterations should have it cached, with the default settings on most distros.

    If you are on Bash, you can use {1..60} instead of $(seq 1 60). Obviously, if you want to use Bash-specific syntax inside the single quotes in the first example, you need bash -c instead of sh -c