bash 如何将结果 /bin/date "%Y-%m-%d %H:%M:%S" 转换为秒？
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How to covert the result /bin/date "%Y-%m-%d %H:%M:%S" into seconds?
提问by marcos barcelona
How can I convert a date variable to seconds?
如何将日期变量转换为秒？
I have a variable called CDATE defined like this:
我有一个名为 CDATE 的变量，定义如下：
CDATE=$(/bin/date +"%Y-%m-%d %H:%M:%S")
CDATE example output: "2012-12-12 12:12:12"
CDATE 示例输出：“2012-12-12 12:12:12”
I would like to convert this to time in seconds and save it in another variable.
我想将其转换为以秒为单位的时间并将其保存在另一个变量中。
回答by Cyrus
x=($(date +"%Y-%m-%d %H:%M:%S %s"))
CDATE="${x[0]} ${x[1]}"
secs="${x[2]}" # seconds since 1970-01-01 00:00:00 UTC
echo $CDATE
echo $secs
Output:
输出：
2015-04-04 14:13:08 1428149588
回答by weibeld
On Linux:
在 Linux 上：
CDATE=$(date "+%Y-%m-%d %H:%M:%S")
SECONDS=$(date -d "$CDATE" +%s)
On Mac:
在 Mac 上：
CDATE=$(date "+%Y-%m-%d %H:%M:%S")
SECONDS=$(date -j -f "%Y-%m-%d %H:%M:%S" "$CDATE" +%s)
And the content of the two variables is (in both cases):
并且两个变量的内容是（在两种情况下）：
$ echo "$CDATE"
2015-04-04 16:24:41
$ echo "$SECONDS"
1428157497
回答by Michael Jaros
If you are not charged for CPU time, then I would call datetwice to make it more readable:
如果您不为 CPU 时间付费，那么我会调用date两次以使其更具可读性：
# get the timestamp:
timestamp=$( date +"%s" )
# format it:
CDATE=$( date -d @"$timestamp" +"%Y-%m-%d %H:%M:%S" )
echo "$CDATE"
echo "$timestamp"
回答by Tiago Lopo
The simplest way:
最简单的方法：
source <(date +"CDATE='%Y-%m-%d %H:%M:%S' EPOCH='%s'")
Example:
例子：
tiago@dell:/tmp$ source <(date +"CDATE='%Y-%m-%d %H:%M:%S' EPOCH='%s'")
tiago@dell:/tmp$ echo "cdate:$CDATE epoch:$EPOCH"
cdate:2015-04-05 12:39:58 epoch:1428233998
