如何在sed中更改日期格式? [英] How to change date format in sed?
问题描述
date
命令将时间戳中的日期转换为另一种格式。在控制台中,我会说 date -d @< timestamp>
,但我恰好想对文本文件中的许多字段执行此操作。 使用 e
在 sed
( sed(GNU sed)4.2.2
,实际上)我在说:
$ echo 1449158360 | sed -r的#。*([0-9] {10})。*#date -d @ \1+%Y;#e'
2015
它的工作,很好!
现在我创建了一个虚拟文件 myfile
:
我的时间戳是1449158360,但我也想知道什么日期是1359199960.
我想替换为相同但具有时间戳的相对年份:
我的时间戳是2015年,但也不知道2013年是哪个日期。
但是,如果我尝试运行与上述相同的命令,它将失败:
$ sed -r's#([0-9] {10})#date -d @\1+%Y;#e'myfile
sh:my:command not found
sh:but:command not found
因为 sed
将第一个单词解释为要执行的内容。
显然,如果我只是获取这些数据,而没有其他的东西,那么它可以工作:
$ sed -r的#。*([0-9] {10})。*#date - d @\1+%Y;#ge'myfile
2015
所以我想知道,在 sed
中捕获的组中,我应该如何调用 date
并替换文本,考虑它被其他文字包围,不得不保持不变?
e
切换到 sed
将 sh -c
替换为不匹配的文本,并从此命令中显而易见:
echo'a 1449158360'| sed -r的#([0-9] {10})#date -d @ \1+%Y;#e'
sh:a:command not found
所以即使我们只匹配 1449158360
但 sh -c
正在运行 a 1449158360
。
由于缺少这个解决方法正则表达式可能会显得很疯狂,但这是您可以如下所示从文件中运行多个匹配输入:
sed -r's#(([^ 0-9] [0-9] {0,9})*)(\ b [0-9] {10} \b)(([0-9] {0,9} [^ 0-9])*)#printf%s%s%s\1$ (日期-d @ \3+%Y)\4;#ge'文件
基本上,我们在这个正则表达式中匹配<之前> 10位数>
strong>输出:
我的时间戳是2015年,但也不知道2013年的日期是多少。
为了澄清所用的正则表达式,我创建了此演示。
绝对不是这个通用解决方案 e
模式问题,将其视为基于正则表达式的解决方法。
Say I want to convert a date in timestamp to another format through the date
command. In console I would say date -d@<timestamp>
but I happen to want to do this to many fields in a text file.
Using the e
to execute in sed
(sed (GNU sed) 4.2.2
, actually) I am saying:
$ echo 1449158360 | sed -r 's#.*([0-9]{10}).*#date -d@\1 "+%Y";#e'
2015
It works, nice!
Now I created a dummy file myfile
:
my timestamp is 1449158360 but also I wonder what date was 1359199960.
Which I would like to have replaced to the same but having the relative year of the timestamps:
my timestamp is 2015 but also I wonder what date was 2013.
However, if I try to run the same command as above it fails:
$ sed -r 's#([0-9]{10})#date -d@"\1" "+%Y";#e' myfile
sh: my: command not found
sh: but: command not found
Because sed
interprets the first words as something to execute.
Obviously it works if I just fetch these data and nothing else:
$ sed -r 's#.*([0-9]{10}).*#date -d@"\1" "+%Y";#ge' myfile
2015
So I wonder: what should I do to call date
against captured groups in sed
and replace text with it, considering it is surrounded by other text that have to remain untouched?
e
switch in sed
substitute applies sh -c
to unmatched text as well as evident from this command:
echo 'a 1449158360' | sed -r 's#([0-9]{10})#date -d@\1 "+%Y";#e'
sh: a: command not found
So even though we are matching only 1449158360
but sh -c
is being run on a 1449158360
.
Due to absence of non-greedy and lookaheads regex in sed
this workaround regex might appear crazy but this is how you can run it for multiple matching input from file as in your question:
sed -r 's#(([^0-9][0-9]{0,9})*)(\b[0-9]{10}\b)(([0-9]{0,9}[^0-9])*)#printf "%s%s%s" "\1" $(date -d@\3 "+%Y") "\4";#ge' file
Basically we are matching <before>10-digits<after>
in this regex.
Output:
my timestamp is 2015 but also I wonder what date was 2013.
To clarify the regex used I have created this demo.
By no means this is a generic solution to e
mode issue, treat it as a regex based workaround.
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