考虑下面的伪代码:
#!/bin/ksh
rangeStartTime_hr=13
rangeStartTime_min=56
rangeEndTime_hr=15
rangeEndTime_min=05
getCurrentMinute() {
return `date +%M | sed -e 's/0*//'`;
# Used sed to remove the padded 0 on the left. On successfully find&replacing
# the first match it returns the resultant string.
# date command does not provide minutes in long integer format, on Solaris.
}
getCurrentHour() {
return `date +%l`; # %l hour ( 1..12)
}
checkIfWithinRange() {
if [[ getCurrentHour -ge $rangeStartTime_hr &&
getCurrentMinute -ge $rangeStartTime_min ]]; then
# Ahead of start time.
if [[ getCurrentHour -le $rangeEndTime_hr &&
getCurrentMinute -le $rangeEndTime_min]]; then
# Within the time range.
return 0;
else
return 1;
fi
else
return 1;
fi
}
有没有更好的实施方式checkIfWithinRange()
?UNIX 中是否有任何内置函数可以更轻松地执行上述操作?我是 korn 脚本的新手,非常感谢您的意见。