unix - 检查当前时间是否在 UNIX 上定义的时间范围内

Question

考虑下面的PSUEDO-CODE:

#!/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
}

是否有更好的方法来实现 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 脚本编写的新手，非常感谢您的意见。

Answer 1

return 命令用于返回退出状态，而不是任意字符串。这与许多其他语言不同。您使用 stdout 传递数据:

getCurrentMinute() {
    date +%M  | sed -e 's/^0//' 
    # make sure sed only removes zero from the beginning of the line
    # in the case of "00" don't be too greedy so only remove one 0
}

此外，您需要更多语法来调用该函数。当前您正在比较 if 条件中的文字字符串 "getCurrentMinute"

if [[ $(getCurrentMinute) -ge $rangeStartTime_min && ...

如果有点不同的话我会做

start=13:56
end=15:05

checkIfWithinRange() {
    current=$(date +%H:%M) # Get's the current time in the format 05:18
    [[ ($start = $current || $start < $current) && ($current = $end || $current < $end) ]] 
}

if checkIfWithinRange; then
    do something
fi