如何使用 bash 中可用的工具生成范围内所有日期的列表？
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How to generate a list of all dates in a range using the tools available in bash?
提问by Hjulle
I want to download a bunch of files named with ISO-8601 dates. Is there a simple way to do this using bash+GNU coreutils? (Or some trick to make wget/curl to generate the list automatically, but I find that unlikely)
我想下载一堆以 ISO-8601 日期命名的文件。有没有一种简单的方法可以使用 bash+GNU coreutils 来做到这一点？（或者一些让 wget/curl 自动生成列表的技巧，但我发现这不太可能）
Similar to this question, but not restricted to weekdays: How to generate a range of nonweekend dates using tools available in bash?. I guess that there is a simpler way to do it without that restriction.
类似于这个问题，但不限于工作日：如何使用 bash 中可用的工具生成一系列非周末日期？. 我想有一种更简单的方法可以不受限制地做到这一点。
Also related to How to generate date range for random data on bash, but not restricted to a single year.
还与How to generate date range for random data on bash 有关，但不限于一年。
回答by Tom Fenech
If you have GNU date, you could do use either a forloop in any POSIX-compliant shell:
如果您有 GNU date，则可以在任何符合 POSIX 的 shell 中使用for循环：
# with "for"
for i in {1..5}; do
echo $(date -I -d "2014-06-28 +$i days")
done
or an untilloop, this time using Bash's extended test [[:
或until循环，这次使用 Bash 的扩展测试[[：
# with "until"
d="2014-06-29"
until [[ $d > 2014-07-03 ]]; do
echo "$d"
d=$(date -I -d "$d + 1 day")
done
Note that non-ancient versions of shwill also do lexicographical comparison if you change the condition to [ "$d" \> 2014-07-03 ].
请注意，sh如果将条件更改为[ "$d" \> 2014-07-03 ].
Output from either of those loops:
这些循环中的任何一个的输出：
2014-06-29
2014-06-30
2014-07-01
2014-07-02
2014-07-03
For a more portable way to do the same thing, you could use a Perl script:
对于做同样事情的更便携的方式，你可以使用 Perl 脚本：
use strict;
use warnings;
use Time::Piece;
use Time::Seconds;
use File::Fetch;
my ($t, $end) = map { Time::Piece->strptime($_, "%Y-%m-%d") } @ARGV;
while ($t <= $end) {
my $url = "http://www.example.com/" . $t->strftime("%F") . ".log";
my $ff = File::Fetch->new( uri => $url );
my $where = $ff->fetch( to => '.' ); # download to current directory
$t += ONE_DAY;
}
Time::Piece, Time::Seconds and File::Fetch are all core modules. Use it like perl wget.pl 2014-06-29 2014-07-03.
Time::Piece、Time::Seconds 和 File::Fetch 都是核心模块。像perl wget.pl 2014-06-29 2014-07-03.
回答by glenn Hymanman
Using GNU date and bash:
使用 GNU 日期和 bash：
start=2014-12-29
end=2015-01-03
while ! [[ $start > $end ]]; do
echo $start
start=$(date -d "$start + 1 day" +%F)
done
2014-12-29
2014-12-30
2014-12-31
2015-01-01
2015-01-02
2015-01-03
回答by Ernst de Haan
If you are on macOS, then dateworks a bit different from GNU date. Here's a variant to Tom Fenech's dateinvocation that supports both GNU and Darwin:
如果您使用的是 macOS，则其date工作方式与 GNU 略有不同date。这是 Tom Fenechdate调用的一个变体，它支持 GNU 和 Darwin：
if [ $(uname) = 'Darwin' ]; then
d=$(date -j -v+1d -f %Y-%m-%d $d +%Y-%m-%d)
elif [ $(uname) = 'Linux' ]; then
d=$(date -I -d "$d + 1 day")
fi
回答by Hjulle
This is how I ended up doing it:
这就是我最终这样做的方式：
d=$(date -I);
while wget "http://www.example.com/$d.log"; do
d=$(date -I -d "$d - 1 day");
done
This tries to download all files from today's date until we get a 404.
这会尝试从今天开始下载所有文件，直到我们得到 404。
回答by runlevel0
I use this handy function to work with log files in the format yyyymmdd.log.gz:
我使用这个方便的函数来处理yyyymmdd.log.gz格式的日志文件：
function datelist { for dt in $(seq -w ) ; do date -d $dt +'%Y%m%d' 2>/dev/null ; done ; }
It accepts dates in the format yyyymmdd.
它接受yyyymmdd格式的日期。
