UNIX日期:如何将星期数(日期+％W)转换为日期范围(星期一至星期日)? [英] UNIX date: How to convert week number (date +%W) to a date range (Mon-Sun)?

问题描述

我有一个从庞大的日志文件中提取的周数列表，它们是使用以下语法提取的:

I have list of week numbers extracted from huge log file, they were extracted using syntax:

$ date --date="Wed Mar 20 10:19:56 2012" +%W;
12

我想创建一个简单的bash函数，可以将这些星期数转换为日期范围.我想函数应该接受2个参数:$ number和$ year，例如:

I want to create a simple bash function which can convert these week numbers to a date range. I suppose function should accept 2 arguments: $number and $year, example:

$ week() { ......... }
$ number=12; year=2012
$ week $number $year
"Mon Mar 19 2012" - "Sun Mar 25 2012"

推荐答案

使用GNU date:

$ cat weekof.sh
function weekof()
{
    local week=$1 year=$2
    local week_num_of_Jan_1 week_day_of_Jan_1
    local first_Mon
    local date_fmt="+%a %b %d %Y"
    local mon sun

    week_num_of_Jan_1=$(date -d $year-01-01 +%W)
    week_day_of_Jan_1=$(date -d $year-01-01 +%u)

    if ((week_num_of_Jan_1)); then
        first_Mon=$year-01-01
    else
        first_Mon=$year-01-$((01 + (7 - week_day_of_Jan_1 + 1) ))
    fi

    mon=$(date -d "$first_Mon +$((week - 1)) week" "$date_fmt")
    sun=$(date -d "$first_Mon +$((week - 1)) week + 6 day" "$date_fmt")
    echo "\"$mon\" - \"$sun\""
}

weekof $1 $2
$ bash weekof.sh 12 2012
"Mon Mar 19 2012" - "Sun Mar 25 2012"
$ bash weekof.sh 1 2018
"Mon Jan 01 2018" - "Sun Jan 07 2018"
$

---

注意:

正如OP所述，周号由date +%W获取.根据GNU日期的手册:

---

NOTE:

As the OP mentions, the week number is got by date +%W. According to GNU date's manual:

%W:一年中的第几周，星期一为一周的第一天(00..53)

%W: week number of year, with Monday as first day of week (00..53)

所以:

1. 每个星期从星期一开始.
2. 如果1月1日是星期一，那么第一周将是第一周.
3. 如果1月1日不是星期一，那么前几天将是第0周，而第1周将从第一个星期一开始.

1. Each week would start from Mon.
2. If Jan 1 is Mon, then the first week will be week #1.
3. If Jan 1 is not Mon, then the first few days will be week #0 and the week #1 starts from the first Mon.

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