问题
I have a shellscript with the following lines:
set -o nounset
set -o errexit
set -o xtrace
if [ "$#" -ne 0 ]
then
echo 'A message'
exit 1
fi
Can somebody explain the commands, in particular the setter and the "$#" portion?
回答1:
$# is the number of arguments passed to the script.
So this
if [ "$#" -ne 0 ]
check ensures that if no arguments are passed then the script exits, which implies that the script expects one or more arguments.
In a simple script called my_script, have this:
#!/bin/bash
echo $#
and run with:
$ ./my_script # prints 0
$ ./my_script a b cde # prints 3
$ ./my_script 1 2 3 4 # prints 4
The set built-in options:
set -o unset (equivalent to set -u): Treats unset variable as error.
set -o errexit (equivalent to set -e): Exits immediately on error.
set -o xtrace (equivalent to set -x): Displays the expanded command. Typically used to to debug shell scripts.
Consider a simple script called opt to demonstrate this:
#!/bin/bash
set -e
set -u
set -x
cmd="ps $$"
${cmd}
echo $var # 'var' is unset. So it's an "error". Since we have
# 'set -o e', the script exits.
echo "won't print this
"
outputs something like:
+ cmd='ps 2885'
+ ps 2885
PID TTY STAT TIME COMMAND
2885 pts/1 S+ 0:00 /bin/bash ./s
./s: line 9: var: unbound variable
The first two lines in the output (starting with +) are due to set -x.
The next two are the result of running the ${cmd}.
The next line is the error, happened as the result of set -u.
You can read more about the set built-in options here.
回答2:
In Bash, $# keeps the number of command line arguments. In your case, the conditional part will fire only when there are some.
I believe very similar question was answered here, second or third answer matching your problem.
来源：https://stackoverflow.com/questions/34370525/linux-shellscript