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Bash：检查多个变量中的一个且仅一个是否相等

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Bash：检查多个变量中的一个且仅一个是否相等

当谈到阅读时间时，我正在尝试正确格式化“和”。我想尝试弄清楚如何仅在最后一个值（无论是小时、分钟还是秒）之前回显“和”，并且仅当有多个值时（需要两个值，即：2 天和5 分钟，或 2 天 12 小时 5 分钟）

我不知道正确的语法，如果 H、M 和 S 中只有一个值大于 0，则 $DAnd="and "。这也会重复几个小时，所以如果只有 M 和 S 的值大于 0，则“$HAnd="and "。在几分钟内，这很容易，因为它只检查 S 是否大于0。

如何在不编写一大堆代码的情况下实现这一目标？

我当前的脚本：

#!/bin/bash
TIME1="08/15/2018 10:30:41"
TIME2="08/30/2018 8:34:40"
SEC1=`date +%s -d "${TIME1}"`
SEC2=`date +%s -d "${TIME2}"`
DIFF=`expr ${SEC2} - ${SEC1}`
CONVERTTIME()
{
  local T=$1
  local D=$((T/60/60/24))
  local H=$((T/60/60%24))
  local M=$((T/60%60))
  local S=$((T%60))
  if [[ $D > 0 ]]; then
    [[ ($H = 0 || $M = 0 || $S = 0) && ($H = 0 && $M = 0 && $S = 0) ]] && DComma="" || DComma=","
    #The problem
    [[ ($H > 0 || $M > 0 || $S > 0) && ($H = 0 || $M = 0 || $S = 0) ]] && DAnd="" || DAnd="and "
    [[ $D = 1 ]] && echo -n "$D day$DComma " || echo -n "$D days$DComma $DAnd"

  fi
  if [[ $H > 0 ]]; then
    [[ ($M = 0 || $S = 0) && ($M = 0 && $S = 0) ]] && HComma="" || HComma=","
    #The problem
    [[ ($M > 0 || $S > 0) && ($M = 0 || $S = 0) ]] && HAnd="" || HAnd="and "
    [[ $H = 1 ]] && echo -n "$H hour$HComma $HAnd" || echo -n "$H hours$HComma $HAnd"
  fi
  if [[ $M > 0 ]]; then
    [[ $S = 0 ]] && MComma="" || MComma=", and "
    [[ $M = 1 ]] && echo -n "$M minute$MComma" || echo -n "$M minutes$MComma"
  fi
  if [[ $S > 0 ]]; then
    [[ $S = 0 ]] && echo -n "$S second" || echo -n "$S seconds"
  fi
  # If no difference in time:
  [[ $D = 0 && $H = 0 && $M = 0 && $S = 0 ]] && echo -n "0 seconds"
  echo
}
echo
echo "TIME DIFFERENCE: $(CONVERTTIME $DIFF)"
echo

答案1

我会把它写成

seconds2text() {     
    local diff=$1
    local words=()
    if (( diff == 0 )); then
        words=("0 seconds")
    else
        local s=$((diff % 60))
        local m=$((diff / 60 % 60))
        local h=$((diff / 60 / 60 % 24))
        local d=$((diff / 60 / 60 / 24))

        (( d > 0 )) && { unit=day;    (( d > 1 )) && unit+=s; words+=("$d $unit"); };
        (( h > 0 )) && { unit=hour;   (( h > 1 )) && unit+=s; words+=("$h $unit"); }
        (( m > 0 )) && { unit=minute; (( m > 1 )) && unit+=s; words+=("$m $unit"); }
        (( s > 0 )) && { unit=second; (( s > 1 )) && unit+=s; words+=("$s $unit"); }
        (( ${#words[@]} > 1 )) && words[-1]="and ${words[-1]}"
    fi
    local IFS=,
    local text="${words[*]}"
    text=${text/,and/ and}
    echo "${text//,/, }"
}

进而

$ for d in 0 1 61 3600 3601 3660 3661 86400 86401 86460 86461 90000 90001 90060 90061 180122; do seconds2text $d; done
0 seconds
1 second
1 minute and 1 second
1 hour
1 hour and 1 second
1 hour and 1 minute
1 hour, 1 minute and 1 second
1 day
1 day and 1 second
1 day and 1 minute
1 day, 1 minute and 1 second
1 day and 1 hour
1 day, 1 hour and 1 second
1 day, 1 hour and 1 minute
1 day, 1 hour, 1 minute and 1 second
2 days, 2 hours, 2 minutes and 2 seconds

笔记：

  • [[ $x > 0 ]]并[[ $x = 0 ]]正在做细绳比较，不数字比较。使用((x > 0))and((x == 0))代替（请注意，$那里不需要 ）。
  • 改掉使用全部大写变量的习惯。把它们留给外壳。有一天你会写完PATH=something然后想知道为什么你的剧本被破坏了。

答案2

和zsh：

#! /bin/zsh -
t1=${1?first date please}
t2=${2?second date please}

zmodload zsh/datetime
strftime -rst1 '%m/%d/%Y %H:%M:%S' "$t1" || exit
strftime -rst2 '%m/%d/%Y %H:%M:%S' "$t2" || exit

t=$((t2 - t1))
if ((t)) {
  and=' and' out= plural=s
  for unit duration (
    second 60
    minute 60
    hour   24
    day    7
    week   t+1
  ) {
    ((n = t % duration))
    ((t /= duration))
    ((n > 0)) && out="$and $n $unit${plural: n<2}$out" and=,
    ((t)) || break
  }
  out=${out#?* }
} else {
  out=now
}
echo "$out"

然后：

./that-script "08/15/2018 10:30:41" "08/30/2018 8:34:40"
2 weeks, 22 hours, 3 minutes and 59 seconds

答案3

如何将date命令与sed脚本结合起来（日期/时间差异小于一年）？尝试

date +"%j d, %H h, %M m, %S s" -d@$(((d-90000))) | sed -r 's/, 0*00 .//g; s/365 d,* *//; s/ ([0-9]{2} [dhms])$/ and \1/; s/(^| )0+/\1/g; s/^$/0 s/; s/([02-9] [dhms])/\1s/g; s/ d(s|,|$)/ day\1/; s/ h/ hour/; s/ m/ minute/; s/ s/ second/'

从 Glenn jackman 的帖子中窃取测试循环：

for DIFF in 0 1 61 3600 3601 3660 3661 86400 86401 86460 86461 90000 90001 90060 90061 180122
  do date +"%j d, %H h, %M m, %S s" -d@$(((DIFF-90000))) |
sed -r '
s/, 0*00 .//g
s/365 d,* *//
s/ ([0-9]{2} [dhms])$/ and \1/
s/(^| )0+/\1/g
s/^$/0 s/
s/([02-9] [dhms])/\1s/g
s/ d(s|,|$)/ day\1/
s/ h/ hour/
s/ m/ minute/
s/ s/ second/
'; done
0 seconds
1 second
1 minute, and 1 second
1 hour
1 hour, and 1 second
1 hour, and 1 minute
1 hour, 1 minute, and 1 second
1 day
1 day, and 1 second
1 day, and 1 minute
1 day, 1 minute, and 1 second
1 day, and 1 hour
1 day, 1 hour, and 1 second
1 day, 1 hour, and 1 minute
1 day, 1 hour, 1 minute, and 1 second
2 days, 2 hours, 2 minutes, and 2 seconds

（我知道它无法s在任何以 1 结尾的数字上添加复数，例如 11、21 等...）

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