使用bash减去时间？

Question

是否可以使用 bash 减去包含 24 小时时间的变量？

#!/bin/bash
var1="23:30" # 11:30pm
var2="20:00" # 08:00pm

echo "$(expr $var1 - $var2)"

运行它会产生以下错误。

./test 
expr: non-integer argument

我需要输出以十进制形式出现，例如：

./test 
3.5

Answer 1

该date命令的输入非常灵活。你可以利用它来发挥你的优势：

#!/bin/bash
var1="23:30"
var2="20:00"

# Convert to epoch time and calculate difference.
difference=$(( $(date -d "$var1" "+%s") - $(date -d "$var2" "+%s") ))

# Divide the difference by 3600 to calculate hours.
echo "scale=2 ; $difference/3600" | bc

输出：

$ ./test.bash
3.50

Answer 2

使用 only bash，没有外部程序，你可以这样做：

#!/bin/bash

# first time is the first argument, or 23:30     
var1=${1:-23:30}
# second time is the second argument, or 20:00
var2=${2:-20:00}

# Split variables on `:` and insert pieces into arrays
IFS=':' read -r -a t1 <<< "$var1"
IFS=':' read -r -a t2 <<< "$var2"

# strip leading zeros (so it's not interpreted as octal
t1=("${t1[@]##0}")
t2=("${t2[@]##0}")

# check if the first time is before the second one
if (( t1[0] > t2[0] || ( t1[0] == t2[0] && t1[1] > t2[1]) ))
then
  # if the minutes on the first time are less than the ones on the second time
  if (( t1[1] < t2[1] ))
  then
    # add 60 minutes to time 1
    (( t1[1] += 60 ))
    # and subtract an hour
    (( t1[0] -- ))
  fi
  # now subtract the hours and the minutes
  echo $((t1[0] -t2[0] )):$((t1[1] - t2[1]))
  # to get a decimal result, multiply the minutes by 100 and divide by 60
  echo $((t1[0] -t2[0] )).$(((t1[1] - t2[1])*100/60))
else
  echo "Time 1 should be after time 2" 2>&1
fi

测试：

$ ./script.sh 
3:30
3.50

$ ./script.sh 12:10 11:30
0:40
0.66

$ ./script.sh 12:00 11:30
0:30
0.50

如果您想要更复杂的时差，可能跨越不同的日期等，那么最好使用 GNU 日期。