Arrays 如何将数组日期添加到datediff计算中

新手在这里学习bash脚本，如果你有时间，可以在这方面提供一些帮助。我的客户上传文件，每个文件名都有一个日期戳，如下所示：

* 20170815041135 
* 20170820041135 
* 20170823071727 
* 20170826040609   
* 20170828050704 
* 20170830153011

我需要计算每次上传之间的天数，然后找到列出的上传的平均间隔

我可以用这个命令找到两个日期之间的日期差

echo $(( ($(date --date="20170831" +'%s' ) - $(date --date="20170821" +'%s')) / (60*60*24) )) 

给10 要执行多个日期，我已经读到我需要一个数组，所以这里是我在一个数组中上传日期的范围

array=( `20170830153011`,`20170828050704`,`20170826040609`,`20170823071727`,`20170820041135`,`20170815041135` )

我读过，我需要这样循环计算

for i in "${array[@]}" do 
  ?

---

如何将数组日期添加到计算中？

将日期时间添加到数组中：

timestamps=(
    20170815041135
    20170820041135
    20170823071727
    20170826040609
    20170828050704
    20170830153011
)

现在让我们将其转换为大纪元：

epochs=()
for timestamp in "${timestamps[@]}"; do 
    iso8601=$(sed -r 's/(....)(..)(..)(..)(..)(..)/\1-\2-\3T\4:\5:\6/' <<<"$timestamp")
    epochs+=( "$(date -d "$iso8601" "+%s")" )
done
printf "%s\n" "${epochs[@]}"

现在我们可以迭代它们来计算差异。请注意，bash数组索引从零开始：

n=0
sum=0
for ((i=1; i < "${#epochs[@]}"; i++ )); do 
    ((n++, diff=(${epochs[i]} - ${epochs[i-1]}), sum+=diff))
    echo "diff $n = $diff seconds = $((diff/86400)) days"
done
echo "average = $((sum/n)) seconds = $((sum/n/86400)) days"

---

从1970年开始以秒为单位转换日期。 计算差额。

---

我希望bash date函数知道从该日期开始考虑夏令时。

“每个文件名中都有一个日期戳”--完整的文件名是什么？XXX_1_20170830200211.bin

filename=XXX_1_20170830200211.bin；tmp=${filename%.bin}；datestamp=${tmp##*.}

--谢谢你，格伦，非常感谢你这样安排

n=0
sum=0
for ((i=1; i < "${#epochs[@]}"; i++ )); do 
    ((n++, diff=(${epochs[i]} - ${epochs[i-1]}), sum+=diff))
    echo "diff $n = $diff seconds = $((diff/86400)) days"
done
echo "average = $((sum/n)) seconds = $((sum/n/86400)) days"

diff 1 = 432000 seconds = 5 days
diff 2 = 270352 seconds = 3 days
diff 3 = 247722 seconds = 2 days
diff 4 = 176455 seconds = 2 days
diff 5 = 210187 seconds = 2 days
average = 267343 seconds = 3 days