在Linux中获取给定日期的前一个工作日的功能

给定输入日期,我想编写一个bash函数,该函数将输出前一个工作日.
我的意思是前一个工作日(星期一至星期五)；
我不需要考虑假期.
因此,例如,给定“ 2018年1月2日”,结果应为“ 2018年1月1日”
(即使那是个假期),
但给定的是“ 2018年1月1日”,则结果应为“ 2017年12月29日”
(因为12月30日和31日分别是星期六和星期日).
我不需要任何特殊格式；
只是迄今为止人类可以理解并且可以接受的东西-d.

我已经尝试了以下方法,但是似乎没有正确考虑输入日期：

function get_prevIoUs_busday()
{
    DAY_OF_WEEK=`$1 +%w`
    if [ $DAY_OF_WEEK -eq 0 ] ; then
        LOOKBACK=-2
    elif [ $DAY_OF_WEEK -eq 1 ] ; then
        LOOKBACK=-3
    else
        LOOKBACK=-1
    fi
    PREVDATE=date -d "$1 $LOOKBACK day"
}

我想今天将其应用：

PREVDATE=$(get_prevIoUs_busday $(date)) 
echo $PREVDATE

而昨天：

PREVDATE=$(get_prevIoUs_busday (date -d "$(date) -1 day")) 
echo $PREVDATE

但它不起作用：

main.sh: line 3: Fri: command not found 
main.sh: line 4: [: -eq: unary operator expected 
main.sh: line 6: [: -eq: unary operator expected 
main.sh: line 11: -d: command not found 
main.sh: command substitution: line 20: Syntax error near unexpected token `date'
main.sh: command substitution: line 20: `get_prevIoUs_busday (date -d "$(date) -1 day"))'

解决方法:

一个您想要做的功能是：

get_prevIoUs_busday() {
        if [ "$1" = "" ]
        then
                printf 'Usage: get_prevIoUs_busday (base_date)\n' >&2
                return 1
        fi
        base_date="$1"
        if ! day_of_week="$(date -d "$base_date" +%u)"
        then
                printf 'Apparently "%s" was not a valid date.\n' "$base_date" >&2
                return 2
        fi
        case "$day_of_week" in
          (0|7)         # Sunday should be 7, but apparently some people
                        # expect it to be 0.
                offset=-2       # Subtract 2 from Sunday to get Friday.
                ;;
          (1)   offset=-3       # Subtract 3 from Monday to get Friday.
                ;;
          (*)   offset=-1       # For all other days, just go back one day.
        esac
        if ! prev_date="$(date -d "$base_date $offset day")"
        then
                printf 'Error calculating $(date -d "%s").\n' "$base_date $offset day"
                return 3
        fi
        printf '%s\n' "$prev_date"
}

例如,

$get_prevIoUs_busday
Usage: get_prevIoUs_date (base_date)
$get_prevIoUs_busday foo
date: invalid date ‘foo’
Apparently "foo" was not a valid date.
$get_prevIoUs_busday today
Fri, Nov 30, 2018  1:52:15 AM
$get_prevIoUs_busday "$(date)"
Fri, Nov 30, 2018  1:52:51 AM
$PREVDATE=$(get_prevIoUs_busday $(date))
$echo "$PREVDATE"
Fri, Nov 30, 2018 12:00:00 AM
$get_prevIoUs_busday "$PREVDATE"
Thu, Nov 29, 2018 12:00:00 AM
$PREVPREVDATE=$(get_prevIoUs_busday "$PREVDATE")
$printf '%s\n' "$PREVPREVDATE"
Thu, Nov 29, 2018 12:00:00 AM
$get_prevIoUs_busday "$PREVPREVDATE"
Wed, Nov 28, 2018 12:00:00 AM