linux - 在 Linux 中获取给定日期的前一个工作日的函数

Question

给定一个输入日期，我想编写一个 bash 函数来输出前一个工作日。我指的是前周的一天(周一到周五)；我不需要它来考虑假期。因此，例如，给定“2018 年 1 月 2 日”，结果应为“2018 年 1 月 1 日”(即使那是假期)，但给定“2018 年 1 月 1 日”，结果应为“2017 年 12 月 29 日”(因为 12 月 30 日和 31 日是周六和周日)。我不需要任何特定的格式；只是人类可读且 date -d 可接受的内容。

我尝试了以下方法，但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"
}

我今天要申请:

PREVDATE=$(get_previous_busday $(date)) 
echo $PREVDATE

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"))'

Answer 1

做你想做的事情的功能是:

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