Question

问： given_day_of_the_week的最新日期怎么样？

示例：例如，如果今天是 2016-04-21 ，我只想获取given_day_of_the_week的日期，例如{{1这将是Monday。如果代码明天运行，直到下一个2016-04-18前一天仍然会返回此日期。

说明：代码应返回最新的given_day_of_the_week（given_day_of_the_week）日期，直到2016-04-24周日，然后运行相同的代码下周返回Monday。

Answer 1

需要GNU日期：

today_dow=$(date +%w)
days=(Sunday Monday Tuesday Wednesday Thursday Friday Saturday)
for ((dow=0; dow<7; dow++)); do 
    if ((dow < today_dow)); then 
        date -d "last ${days[dow]}"
    else 
        date -d "${days[dow]}"
    fi
done

Sun Apr 17 00:00:00 EDT 2016
Mon Apr 18 00:00:00 EDT 2016
Tue Apr 19 00:00:00 EDT 2016
Wed Apr 20 00:00:00 EDT 2016
Thu Apr 21 00:00:00 EDT 2016
Fri Apr 22 00:00:00 EDT 2016
Sat Apr 23 00:00:00 EDT 2016

所以我们可以这样做：

given_day_of_the_week() {
    local days=(Sunday Monday Tuesday Wednesday Thursday Friday Saturday)
    local today_dow=$(date +%w)
    local dow datestr
    for ((dow=0; dow<7; dow++)); do 
        if [[ "${days[dow],,}" == "${1,,}" ]]; then
            if ((dow < today_dow)); then 
                datestr="last ${days[dow]}"
            else
                datestr="${days[dow]}"
            fi
            date -d "$datestr" "+%F"
        fi
    done
}

导致：

$ given_day_of_the_week tuesday
2016-04-19
$ given_day_of_the_week friday
2016-04-22

如果您使用其他语言环境，那么硬编码这样的工作日名称会给您带来问题

回应@ ryenus的评论：

$ given_day_of_the_week() { 
    local -A days=([sunday]=0 [monday]=1 [tuesday]=2 [wednesday]=3 [thursday]=4 [friday]=5 [saturday]=6)
    local today_dow=$(date +%w)
    local datestr=${1,,}
    local dow=${days[$datestr]}
    [[ -z "$dow" ]] && { echo "error: unknown day: '$1'" >&2; return 1; }
    (( dow < today_dow )) && datestr="last $datestr"
    date -d "$datestr" "+%F"
}
$ given_day_of_the_week friday
2016-04-22
$ given_day_of_the_week monday
2016-04-18
$ given_day_of_the_week FOO
error: unknown day: 'FOO'

Answer 2

受@ glenn-jackman的回答启发，但更简单：

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            $(window).bind('beforeunload', function () {
                if (unsaved) {
                    return "You have made some changes. Save your pending changes before existing the page";
                }
            });
            // Monitor dynamic inputs
            $(document).on('change', ':input', function () { 
                unsaved = true;
            });          
        })

注意：如果在星期天使用，这个答案会给出明天的星期一日期。如果在星期一使用，它还会给出明天的星期二日期。不是100％确定这是你的想法。

如果相反的想法是总是与当前的星期日 - 星期六一起去，这将会这样做：

given_day_of_the_week=$1
f='%A %Y-%m-%d'
tomorrow=`date -d 'tomorrow' +"$f"`
today=`date -d 'today' +"$f"`
last5=`seq 1 5 | xargs -I{} date -d '{} day ago' +"$f"`
echo -e "$tomorrow\n$today\n$last5" |
    grep -i $given_day_of_the_week |
    cut -d' ' -f2

Answer 3

# requires bash 4.x and GNU date
last_kday() {
  local kday=$1
  local -A numbers=([sunday]=0   [monday]=1 [tuesday]=2 [wednesday]=3
                    [thursday]=4 [friday]=5 [saturday]=6)
  if [[ $kday == *day ]]; then
    kday=${numbers[${kday,,}]}
  elif [[ $kday != [0-6] ]]; then
    echo >&2 "Usage: last_kday weekday"
    return 1
  fi

  local today=$(date +%w)
  local days_ago=$(( today - kday ))
  if (( days_ago < 0 )); then let days_ago+=7; fi
  date -d "$days_ago days ago" +%F
}

如何在Bash中获取一周中特定日期的最新日期

3 个答案: