Linux 在 bash 中转换日期格式
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Convert date formats in bash
提问by vehomzzz
I have a date in this format: "27 JUN 2011" and I want to convert it to 20110627
我有一个这种格式的日期：“27 JUN 2011”，我想将它转换为 20110627
Is it possible to do in bash?
可以用 bash 做吗？
采纳答案by matchew
#since this was yesterday
date -dyesterday +%Y%m%d
#more precise, and more recommended
date -d'27 JUN 2011' +%Y%m%d
#assuming this is similar to yesterdays `date` question from you
#http://stackoverflow.com/q/6497525/638649
date -d'last-monday' +%Y%m%d
#going on @seth's comment you could do this
DATE="27 jun 2011"; date -d"$DATE" +%Y%m%d
#or a method to read it from stdin
read -p " Get date >> " DATE; printf " AS YYYYMMDD format >> %s" `date
-d"$DATE" +%Y%m%d`
#which then outputs the following:
#Get date >> 27 june 2011
#AS YYYYMMDD format >> 20110627
#if you really want to use awk
echo "27 june 2011" | awk '{print "date -d\""FSFS"\" +%Y%m%d"}' | bash
#note | bash just redirects awk's output to the shell to be executed
#FS is field separator, in this case you can use date -d "25 JUN 2011" +%Y%m%d
to print the line
#But this is useful if you have more than one date on a line
note this only works on GNU date
请注意，这仅适用于 GNU 日期
I have read that:
我读过：
Solaris version of date, which is unable to support
-dcan be resolve with replacing sunfreeware.com version of date
Solaris版本的date，无法支持
-d可以通过更换sunfreeware.com版本的date解决
回答by Seth Robertson
20110625
outputs
产出
convert_date () {
local months=( JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC )
local i
for (( i=0; i<11; i++ )); do
[[ = ${months[$i]} ]] && break
done
printf "%4d%02d%02d\n" $(( i+1 ))
}
回答by glenn Hymanman
Just with bash:
只需使用 bash：
d=$( convert_date 27 JUN 2011 )
And invoke it like this
并像这样调用它
d_old="27 JUN 2011"
d=$( convert_date $d_old ) # not quoted
Or if the "old" date string is stored in a variable
或者如果“旧”日期字符串存储在变量中
$ date +"%Y%m%d"
20150330
回答by pjammer
Maybe something changed since 2011 but this worked for me:
也许自 2011 年以来发生了一些变化，但这对我有用：
$ date -j -f "%m/%d/%y %H:%M:%S %p" "8/22/15 8:15:00 am" +"%m%d%y"
082215
No need for the -dto get the same appearing result.
无需-d获得相同的出现结果。
回答by DustinB
On OSX, I'm using -f to specify the input format, -j to not attempt to set any date, and an output format specifier. For example:
在 OSX 上，我使用 -f 来指定输入格式，使用 -j 不尝试设置任何日期，以及一个输出格式说明符。例如：
$ date -j -f "%d %b %Y" "27 JUN 2011" +%Y%m%d
20110627
Your example:
你的例子：
#
# Convert one date format to another
#
# Usage: convert_date_format <input_format> <date> <output_format>
#
# Example: convert_date_format '%b %d %T %Y %Z' 'Dec 10 17:30:05 2017 GMT' '%Y-%m-%d'
convert_date_format() {
local INPUT_FORMAT=""
local INPUT_DATE=""
local OUTPUT_FORMAT=""
local UNAME=$(uname)
if [[ "$UNAME" == "Darwin" ]]; then
# Mac OS X
date -j -f "$INPUT_FORMAT" "$INPUT_DATE" +"$OUTPUT_FORMAT"
elif [[ "$UNAME" == "Linux" ]]; then
# Linux
date -d "$INPUT_DATE" +"$OUTPUT_FORMAT"
else
# Unsupported system
echo "Unsupported system"
fi
}
# Example: 'Dec 10 17:30:05 2017 GMT' => '2017-12-10'
convert_date_format '%b %d %T %Y %Z' 'Dec 10 17:30:05 2017 GMT' '%Y-%m-%d'
回答by David Granqvist
If you would like a bash function that works both on Mac OS X and Linux:
如果您想要一个同时适用于 Mac OS X 和 Linux 的 bash 函数：
data=`date`
datatime=`date -d "${data}" '+%Y%m%d'`
echo $datatime
20190206
回答by Matt Vegas
It's enough to do:
这样做就足够了：
data=`date`
datatime=`date -d "${data}" '+%Y%m%d %T'`
echo $data
Wed Feb 6 03:57:15 EST 2019
echo $datatime
20190206 03:57:15
If you want to add also the time you can use in that way
如果您还想添加可以以这种方式使用的时间
##代码##