BASH:如何查找没有。两个日期之间的天数(仅考虑 "Network / Business Days")(即不包括周末 Saturday/Sunday)

BASH: How to find no. of days (considering only "Network / Business Days") between two dates (i.e. exclude weekends Saturday/Sunday)

RHEL7.5

BASH GNU bash,版本 4.2.46(2)-release (x86_64-redhat-linux-gnu)

MS Excel中,我可以使用网络天数来查找号码。两个日期之间的天数。想知道是否可以使用 bash 作为第一偏好(--或 Linux 上的任何其他 pre-installed 语言支持使用 one-liner 来解决这个问题——我的第二偏好)。我不确定 Linux 中是否存在计算此值的库或自定义 tool/utility。

我有一个 file.txt 包含 YYYY-mm-dd 格式的 2 个列字段,用于 Resolved开始 日期(您现在可以忽略 header 行):

Resolved,StartOfWork
2020-01-16,2020-01-10
2020-01-13,2020-01-13
2020-01-20,2020-01-15
2020-01-20,2020-01-14
2020-01-14,2020-01-09
2020-01-09,2020-01-08
2020-01-16,2020-01-14
2020-01-09,2020-01-07
2020-01-14,2020-01-12

对于每一行,我想计算 否。网络,即这两个日期之间仅工作日如果Resolved/StartOfWork日期在周末则无关紧要:Saturday/Sunday)。

PS:就此 post 的目的而言,我的问题与此不同 post 要求:How to find the difference in days between two dates?

我会为此打电话给 Python 口译员。采用来自 Using Python to count the number of business days in a month? --

的公认答案
countBusinessDaysPy=$(cat <<'EOF'
import datetime, sys

businessdays = 0
startDate = datetime.date.fromisoformat(sys.argv[1])
endDate = datetime.date.fromisoformat(sys.argv[2])
if endDate < startDate:
    (startDate, endDate) = (endDate, startDate)

while startDate <= endDate:      # change from <= to < to not count both start and end days
    if startDate.weekday() < 5:
        businessdays += 1
    startDate += datetime.timedelta(days=1)

print(businessdays)
EOF
)

countBusinessDays() { python3 -c "$countBusinessDaysPy" "$@"; }

...给你一个 shell 函数调用 Python 解释器来做你需要的数学(注意这是一个 inclusive 范围).此后:

$ countBusinessDays 2019-01-01 2020-01-01
262
$ countBusinessDays 2019-01-01 2019-01-07
5

在您的文件上调用此循环(请注意,在现实世界中,我会在 Python 中进行循环,而不是在 bash 中进行循环)可能看起来像:

{
  read -r header; printf '%s\n' "$header,TotalDates"
  while IFS=, read -r resolved startOfWork rest; do
    printf '%s\n' "${resolved},${startOfWork}${rest:+,$rest},$(countBusinessDays "$startOfWork" "$resolved")"
  done
} <yourInputFile

...作为输出发出:

Resolved,StartOfWork,TotalDates
2020-01-16,2020-01-10,5
2020-01-13,2020-01-13,1
2020-01-20,2020-01-15,4
2020-01-20,2020-01-14,5
2020-01-14,2020-01-09,4
2020-01-09,2020-01-08,2
2020-01-16,2020-01-14,3
2020-01-09,2020-01-07,3
2020-01-14,2020-01-12,2

这可能是重新发明轮子,但这里有一个bash解决方案(如果有兴趣的话)。
请注意,它需要 date 命令的 -d 选项。

while IFS="," read -r endday startday; do
    if (( lineno++ == 0 )); then                # handle header line
        echo "Resolved,StartOfWork,TotalDates"
        continue
    fi
    startsec=$(date -d "$startday" +%s)
    startdayofweek=$(date -d "$startday" +%w)   # 0 for Sun, ... 6 for Sat
    endsec=$(date -d "$endday" +%s)
    days=$(( (endsec - startsec) / 86400 + 1 )) # calendar days
    weeks=$(( days / 7 ))                       # number of weeks
    frac=$(( days % 7 ))                        # fraction mod 7
    if (( startdayofweek == 0 )); then          # case of starting on Sunday
        if (( frac > 0 )); then
            add=1                               # additional number of holidays
        else
            add=0
        fi
    else
        magic=$(( frac + (startdayofweek + 6) % 7 ))
                                                # calculate number of holidays
                                                # in the fraction period
        if (( magic < 6 )); then
            add=0
        elif (( magic == 6 )); then
            add=1
        else
            add=2
        fi
    fi
    holidays=$(( weeks * 2 + add ))             # total number of holidays
    workdays=$(( days - holidays ))             # subtract the holidays
    echo "$endday,$startday,$workdays"
done < inputfile