英文:
How to convert normal text into date format in bash shell?
问题
将其从
格式1：12272019（这只是一个数字）
转换为
格式2：2019-12-27
我尝试使用以下方法：
date -d '12272019' +%Y-%m-%d
但它显示无效的日期格式。
英文:
I want to convert it from
Format 1: 12272019(this is just a number)
To
Format 2: 2019-12-27
I tried using below:
date -d '12272019' +%Y-%m-%d
But it is showing invalid date format
答案1
得分: 3
The date command only accepts a predefined set of formats for its input. Your format mmddyyyy is not part of that set.
You can re-arrange your date string manually using either sed:
sed -E 's/(..)(..)(....)/--/' <<< 12272019
or bash:
date=12272019
echo "${date:4}-${date:0:2}-${date:2:2}"
英文:
The date command only accepts a predefined set of formats for its input. Your format mmddyyyy is not part of that set.
You can re-arrange your date string manually using either sed:
sed -E 's/(..)(..)(....)/--/' <<< 12272019
or bash:
date=12272019
echo "${date:4}-${date:0:2}-${date:2:2}"
答案2
得分: 0
在OS/X上，date有一些不同之处，因此您可以为date指定输入格式：
date -jf "%m%d%Y" 12272019 +"%Y-%m-%d"
# => 2019-12-27
然而，Linux不允许您这样做，但您可以使用sed轻松实现：
echo 12272019 | sed -e 's/\(..\)\(..\)\(....\)/--/'
# => 2019-12-27
英文:
On OS/X, date is quite a bit different, so you can specify the input format for date:
date -jf "%m%d%Y" 12272019 +"%Y-%m-%d"
# => 2019-12-27
Linux does not allow you to do that though, but you can do it easily with sed:
echo 12272019 | sed -e 's/\(..\)\(..\)\(....\)/--/'
# => 2019-12-27
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论