[英]How to use getopts in any random order with long optional or non-optional argument first in bash?
I have a script like 我有一个像
./sample.sh --add employeename -f firstname -a age -d detail
./sample.sh --add employeename -a age -d detail -f firstname
./sample.sh --add employeename -d detail -f firstname
./sample.sh --del employeename
./sample.sh --update employeename -f firstname [age,detail are optional]
here,how could i use getopts for --add, --del, --update and i don't want to use getopt. 在这里,我怎么能对--add,--del,--update使用getopts而我不想使用getopt。 Good suggestion will be appreciated.
好的建议将不胜感激。
Those are really commands, not options, so should be regular positional arguments: 这些实际上是命令,而不是选项,因此应该是常规的位置参数:
./sample.sh add employeename -f firstname -a age -d detail
./sample.sh add employeename -a age -d detail -f firstname
./sample.sh add employeename -d detail -f firstname
./sample.sh del employeename
./sample.sh update employeename -f firstname [age,detail are optional]
which should come first; 应该先到 then the remaining options can be parsed using a command-specific set of options.
然后可以使用特定于命令的选项集来解析其余选项。
cmd=$1
shift
case $cmd in
add) do_add "$@" ;;
del) do_del "$@" ;;
update) do_update "$@" ;;
*) echo "Unrecognized command: $cmd"
exit 1
;;
esac
do_add () {
name=$1
shift
while getopts "f:a:d:" opt "$@"; do
...
done
...
}
do_del () {
...
}
# etc
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