解析日期和时间格式 - Bash
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/26753757/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Parsing date and time format - Bash
提问by onur
I have date and time format like this(yearmonthday):
我有这样的日期和时间格式（年月日）：
20141105 11:30:00
I need assignment year, month, day, hour and minute values to variable.
我需要将年、月、日、小时和分钟值分配给变量。
I can do it year, day and hour like this:
我可以像这样在年、日和小时内做到这一点：
year=$(awk '{print }' log.log | sed 's/^\(....\).*//')
day=$(awk '{print }' log.log | sed 's/^.*\(..\).*//')
hour=$(awk '{print }' log.log | sed 's/^\(..\).*//')
How can I do this for month and minute?
我怎样才能在一个月和一分钟内做到这一点？
--
——
And I need that every line of my log file:
我需要我的日志文件的每一行：
20141105 11:30:00 /bla/text.1
20141105 11:35:00 /bla/text.2
20141105 11:40:00 /bla/text.3
....
I'm trying read line by line this log file and do this:
我正在尝试逐行读取此日志文件并执行以下操作：
mkdir -p "/bla/backup/$year/$month/$day/$hour/$minute"
mv $file "/bla/backup/$year/$month/$day/$hour/$minute"
Here is my not working code:
这是我不工作的代码：
#!/bin/bash
LOG=/var/log/LOG
while read line
do
year=${line:0:4}
month=${line:4:2}
day=${line:6:2}
hour=${line:9:2}
minute=${line:12:2}
file=$(awk '{print }')
if [ -f "$file" ]; then
printf -v path "%s/%s/%s/%s/%s" $year $month $day $hour $minute
mkdir -p "/bla/backup/$path"
mv $file "/bla/backup/$path"
fi
done < $LOG
采纳答案by glenn Hymanman
You don't need to call out to awk to date at all, use bash's substring operations
你根本不需要调用 awk 到 date，使用 bash 的 substring 操作
d="20141105 11:30:00"
yr=${d:0:4}
mo=${d:4:2}
dy=${d:6:2}
hr=${d:9:2}
mi=${d:12:2}
printf -v dir "/bla/%s/%s/%s/%s/%s/\n" $yr $mo $dy $hr $mi
echo "$dir"
/bla/2014/11/05/11/30/
Or directly, without all the variables.
或者直接，没有所有的变量。
printf -v dir "/bla/%s/%s/%s/%s/%s/\n" ${d:0:4} ${d:4:2} ${d:6:2} ${d:9:2} ${d:12:2}
Given your log file:
鉴于您的日志文件：
while read -r date time file; do
d="$date $time"
printf -v dir "/bla/%s/%s/%s/%s/%s/\n" ${d:0:4} ${d:4:2} ${d:6:2} ${d:9:2} ${d:12:2}
mkdir -p "$dir"
mv "$file" "$dir"
done < filename
or, making a big assumption that there are no whitespace or globbing characters in your filenames:
或者，假设文件名中没有空格或通配符：
sed -r 's#(....)(..)(..) (..):(..):.. (.*)#mv /blah/////#' | sh
回答by NeronLeVelu
eval "$(
echo '20141105 11:30:00' \
| sed 'G;s/\(....\)\(..\)\(..\) \(..\):\(..\):\(..\) *\(.\)/Year=Month=Day=Hour=Min=Sec=/'
)"
pass via a assignation string to evaluate. You could easily adapt to also check the content by replacing dot per more specific pattern like [0-5][0-9]for min and sec, ...
通过分配字符串进行评估。您可以轻松地通过替换每个更具体的模式（例如[0-5][0-9]分钟和秒）的点来检查内容，...
posix version so --posixon GNU sed
posix 版本--posix等 GNU sed
回答by Leslie Satenstein
I wrote a function that I usually cut and paste into my script files
我写了一个函数，我通常将它剪切并粘贴到我的脚本文件中
function getdate()
{
local a
a=(`date "+%Y %m %d %H %M %S" | sed -e 's/ / /'`)
year=${a[0]}
month=${a[1]}
day=${a[2]}
hour=${a[3]}
minute=${a[4]}
sec=${a[5]}
}
in the script file, on a line of it's own
在脚本文件中，在它自己的一行上
getdate
获取日期
echo "year=$year,month=$month,day=$day,hour=$hour,minute=$minute,second=$sec"
echo "year=$year,month=$month,day=$day,hour=$hour,minute=$minute,second=$sec"
Of course, you can modify what I provided or use answer [6] above. The function takes no arguments.
当然，您可以修改我提供的内容或使用上面的答案 [6]。该函数不接受任何参数。
回答by Kalanidhi
datecommand also do this work
date命令也做这个工作
#!/bin/bash
year=$(date +'%Y' -d'20141105 11:30:00')
day=$(date +'%d' -d'20141105 11:30:00')
month=$(date +'%m' -d'20141105 11:30:00')
minutes=$(date +'%M' -d'20141105 11:30:00')
echo "$year---$day---$month---$minutes"
回答by Jotne
You can use only one awk
您只能使用一个 awk
month=$(awk '{print substr(,5,2)}' log.log)
year=$(awk '{print substr(,0,4)}' log.log)
minute=$(awk '{print substr(,4,2)}' log.log)
etc
回答by Kent
I guess you are processing the log file, which each line starts with the date string. You may have already written a loop to handle each line, in your loop, you could do:
我猜您正在处理日志文件，其中每一行都以日期字符串开头。您可能已经编写了一个循环来处理每一行，在您的循环中，您可以执行以下操作：
d="$(awk '{print ,}' <<<"$line")"
year=$(date -d"$d" +%Y)
month=$(date -d"$d" +%m)
day=$(date -d"$d" +%d)
min=$(date -d"$d" +%M)
回答by gniourf_gniourf
Don't repeat yourself.
不要重复自己。
d='20141105 11:30:00'
IFS=' ' read -r year month day min < <(date -d"$d" '+%Y %d %m %M')
echo "year: $year"
echo "month: $month"
echo "day: $day"
echo "min: $min"
The trick is to ask dateto output the fields you want, separated by a character (here a space), to put this character in IFSand ask readto do the splitting for you. Like so, you're only executing dateonce and only spawn one subshell.
诀窍是要求date输出您想要的字段，由一个字符（这里是一个空格）分隔，将这个字符放入IFS并要求read为您进行拆分。像这样，您只执行date一次并且只生成一个子shell。
If the date comes from the first line of the file log.log, here's how you can assign it to the variable d:
如果日期来自文件的第一行，以下是log.log将其分配给变量的方法d：
IFS= read -r d < log.log
