我需要用逗号替换空格,然后从字符串中删除某些额外的特定字符。
echo "$d"
>>Mon Apr 22 05:06:00 UTC 2019
jent=$(echo $jt1 | sed 's/[[:space:]]/,/g')
echo "$jent"
>>Mon,Apr,22,05:06:00,UTC,2019 #this does the first job of replacing the space with comma
但我又想删除 UTC 部分和前面的逗号,我该如何实现呢?
期望的输出应该是
Mon,Apr,22,05:06:00,2019
答案1
您可以通过一次 sed 调用来完成此操作,例如
$ echo "Mon Apr 22 05:06:00 UTC 2019" | sed 's/ \(UTC \)\?/,/g'
Mon,Apr,22,05:06:00,2019
答案2
由于您的日期具有标准输出和固定的列数,因此您可以使用 awk 打印所需的数据
[user@server1 ~]$ echo "Mon Apr 22 05:06:00 UTC 2019" | awk '{print $1","$2","$3","$4","$6}'
Mon,Apr,22,05:06:00,2019
答案3
你可以简单地说:
~]$ export jent="$(echo $jent | sed 's/,UTC//g')"
~]$ echo $jent
Mon,Apr,22,05:06:00,2019
答案4
我已经使用 awk 和 sed 通过以下方法完成了。
命令:
echo "Mon Apr 22 05:06:00 UTC 2019"|awk '{$(NF-1)="";print $0}'| sed -r "s/\s+/ /g"| sed "s/ /,/g"
输出:
Mon,Apr,22,05:06:00,2019