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

痴心易碎 提交于 2021-02-07 03:21:40

问题


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'

回答1:


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




回答2:


for is an internal function (not to be confused with functions) of a shell that's why you can't call it. You should explicitly call the binary of the shell that runs with the code like this:

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

With bash:

sudo bash -c 'for n in {1..60}; do echo "$n"; sleep 1; sync; done'
sudo bash -c 'for ((n = 1; n <= 60; ++n)); do echo "$n"; sleep 1; sync; done'



回答3:


It's because the fist semicolon terminates the sudo command, which will make do a new command. The easiest way to fix this is to put the loop inside a file and execute it, like

sudo /bin/bash ./myfile


来源:https://stackoverflow.com/questions/25175596/syntax-error-near-unexpected-token-do-when-run-with-sudo

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