From here: http://www.sat.dundee.ac.uk/psc/watchdog/watchdog-testing.html
在这里:http://www.sat.dundee.ac.uk/psc/watchdog/watchdog-testing.html
for n in $(seq 1 60); do echo $n; sleep 1; sync; done
n元(seq 1 60);echo $ n;睡眠1;同步;完成
I get:
我得到:
:~$ sudo for n in $(seq 1 60); do echo $n; sleep 1; sync; done
bash: syntax error near unexpected token `do'
3 个解决方案
#1
7
The shell parses the command line and because for looks like an argument to sudo, you basically get a do without a for.
shell解析命令行，因为for看起来像sudo的参数，所以基本上不需要for。
To fix it, run the loop in a subshell, either as a separate script, or like this;
要修复它，可以在子shell中运行循环，或者作为单独的脚本，或者像这样;
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:
更好的是，避免运行任何不必要的特权用户:
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.
第一个sudo将需要一个密码，但是后续的迭代应该缓存它，在大多数发行版上使用默认设置。
If you are on Bash, you can use {1..60} instead of $(seq 1 60). Obviously, if you want to use Bash-specific syntax inside the single quotes in the first example, you need bash -c instead of sh -c
如果您在Bash中，可以使用{1.. .60}而不是$(seq1 60)。显然，如果希望在第一个示例中的单引号中使用特定于bash的语法，需要bash -c而不是sh -c
#2
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:
因为是壳层的内部函数(不要和函数混淆)，所以不能调用它。您应该显式地调用shell的二进制，该shell使用如下代码运行:
sudo sh -c 'for n in $(seq 1 60); do echo "$n"; sleep 1; sync; done'
With bash:
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'
#3
1
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
这是因为第一个分号终止sudo命令，该命令将执行一个新命令。解决这个问题的最简单方法是将循环放在文件中并执行它，比如
sudo /bin/bash ./myfile
#1
7
The shell parses the command line and because for looks like an argument to sudo, you basically get a do without a for.
shell解析命令行，因为for看起来像sudo的参数，所以基本上不需要for。
To fix it, run the loop in a subshell, either as a separate script, or like this;
要修复它，可以在子shell中运行循环，或者作为单独的脚本，或者像这样;
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:
更好的是，避免运行任何不必要的特权用户:
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.
第一个sudo将需要一个密码，但是后续的迭代应该缓存它，在大多数发行版上使用默认设置。
If you are on Bash, you can use {1..60} instead of $(seq 1 60). Obviously, if you want to use Bash-specific syntax inside the single quotes in the first example, you need bash -c instead of sh -c
如果您在Bash中，可以使用{1.. .60}而不是$(seq1 60)。显然，如果希望在第一个示例中的单引号中使用特定于bash的语法，需要bash -c而不是sh -c
#2
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:
因为是壳层的内部函数(不要和函数混淆)，所以不能调用它。您应该显式地调用shell的二进制，该shell使用如下代码运行:
sudo sh -c 'for n in $(seq 1 60); do echo "$n"; sleep 1; sync; done'
With bash:
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'
#3
1
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
这是因为第一个分号终止sudo命令，该命令将执行一个新命令。解决这个问题的最简单方法是将循环放在文件中并执行它，比如
sudo /bin/bash ./myfile