Linux 如何比较两个 DateTime 字符串并以小时为单位返回差异？(bash 外壳)
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How to compare two DateTime strings and return difference in hours? (bash shell)
提问by denormalizer
I can do that in php with the following code:
我可以使用以下代码在 php 中做到这一点：
$dt1 = '2011-11-11 11:11:11';
$t1 = strtotime($dt1);
$dt2 = date('Y-m-d H:00:00');
$t2 = strtotime($dt2);
$tDiff = $t2 - $t1;
$hDiff = round($tDiff/3600);
$hDiffwill give me the result in hours.
$hDiff将在数小时内给我结果。
How do I implement the above in bash shell?
如何在 bash shell 中实现上述内容？
采纳答案by another.anon.coward
You could use datecommand to achieve this. man datewill provide you with more details. A bash script could be something on these lines (seems to work fine on Ubuntu 10.04 bash 4.1.5):
您可以使用date命令来实现这一点。man date将为您提供更多详细信息。bash 脚本可能是这些行（似乎在 Ubuntu 10.04 bash 4.1.5 上工作正常）：
#!/bin/bash
# Date 1
dt1="2011-11-11 11:11:11"
# Compute the seconds since epoch for date 1
t1=`date --date="$dt1" +%s`
# Date 2 : Current date
dt2=`date +%Y-%m-%d\ %H:%M:%S`
# Compute the seconds since epoch for date 2
t2=`date --date="$dt2" +%s`
# Compute the difference in dates in seconds
let "tDiff=$t2-$t1"
# Compute the approximate hour difference
let "hDiff=$tDiff/3600"
echo "Approx hour diff b/w $dt1 & $dt2 = $hDiff"
Hope this helps!
希望这可以帮助！
