bash 两个给定日期之间的bash循环
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bash loop between two given dates
提问by Chasester
I'm trying to create a script that will loop through files that have their filenames written in the following format: yyyymmdd.hh.filename.
我正在尝试创建一个脚本，该脚本将遍历以以下格式编写文件名的文件：yyyymmdd.hh.filename。
The script is called with:
该脚本被调用：
./loopscript.sh 20091026.00 23
./loopscript.sh 20091026.11 15
./loopscript.sh 20091026.09 20091027.17
The need is for the script to check each hour between those two given dates/hours.
脚本需要检查这两个给定日期/小时之间的每个小时。
e.g.
例如
cat 20091026.00.filename |more
cat 20091026.01.filename |more
...
cat 20091026.23.filename |more
cat 20091027.01.filename |more
cat 20091027.02.filename |more
...
and so on.
等等。
any idea how to go about this? I don't have any difficulty with standard 0 - x loops. or simple for loops. Just not sure how to go about the above.
知道怎么做吗？我对标准的 0 - x 循环没有任何困难。或简单的 for 循环。只是不知道如何解决上述问题。
采纳答案by paxdiablo
To process each file between two given date/hours, you can use the following:
要处理两个给定日期/小时之间的每个文件，您可以使用以下命令：
#!/usr/bin/bash
#set -x
usage() {
echo 'Usage: loopscript.sh <from> <to>'
echo ' <from> MUST be yyyymmdd.hh or empty, meaning 00000000.00'
echo ' <to> can be shorter and is affected by <from>'
echo ' e.g., 20091026.00 27.01 becomes'
echo ' 20091026.00 20091027.01'
echo ' If empty, it is set to 99999999.99'
echo 'Arguments were:'
echo " '${from}'"
echo " '${to}'"
}
# Check parameters.
from="00000000.00"
to="99999999.99"
if [[ ! -z "" ]] ; then
from=
fi
if [[ ! -z "" ]] ; then
to=
fi
## Insert this to default to rest-of-day when first argument
## but no second argument. Basically just sets second
## argument to 23 so it will be transformed to end-of-day.
#if [[ ! -z ""]] ; then
# if [[ -z ""]] ; then
# to=23
# fi
#fi
if [[ ${#from} -ne 11 || ${#to} -gt 11 ]] ; then
usage
exit 1
fi
# Sneaky code to modify a short "to" based on the start of "from".
# ${#from} is the length of ${from}.
# $((${#from}-${#to})) is the length difference between ${from} and ${to}
# ${from:0:$((${#from}-${#to}))} is the start of ${from} long enough
# to make ${to} the same length.
# ${from:0:$((${#from}-${#to}))}${to} is that with ${to} appended.
# Voila! Easy, no?
if [[ ${#to} -lt ${#from} ]] ; then
to=${from:0:$((${#from}-${#to}))}${to}
fi
# Process all files, checking that they're inside the range.
echo "From ${from} to ${to}"
for file in [0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9].[0-9][0-9].* ; do
if [[ ! ( ${file:0:11} < ${from} || ${file:0:11} > ${to} ) ]] ; then
echo " ${file}"
fi
done
When you create the files 20091026.00.${RANDOM}through 20091028.23.${RANDOM}inclusive, this is a couple of sample runs:
当您20091026.00.${RANDOM}通过20091028.23.${RANDOM}包含创建文件时，这是几个示例运行：
pax> ./loopscript.sh 20091026.07 9
From 20091026.07 to 20091026.09
20091026.07.21772
20091026.08.31390
20091026.09.9214
pax> ./loopscript.sh 20091027.21 28.02
From 20091027.21 to 20091028.02
20091027.21.22582
20091027.22.30063
20091027.23.29437
20091028.00.14744
20091028.01.6827
20091028.02.10366
pax> ./loopscript.sh 00000000.00 99999999.99 # or just leave off the parameters.
20091026.00.25772
20091026.01.25964
20091026.02.21132
20091026.03.3116
20091026.04.6271
20091026.05.14870
20091026.06.28826
: : :
20091028.17.20089
20091028.18.13816
20091028.19.7650
20091028.20.20927
20091028.21.13248
20091028.22.9125
20091028.23.7870
As you can see, the first argument must be of the correct format yyyymmdd.hh. The second argument can be shorter since it inherits the start of the first argument to make it the correct length.
如您所见，第一个参数必须具有正确的格式yyyymmdd.hh。第二个参数可以更短，因为它继承了第一个参数的开头以使其长度正确。
This only attempts to process files that exist (from ls) and of the correct format, notevery date/hour within the range. This will be more efficient if you have sparse files (including at the start and the end of the range) since it doesn't need to check that the files exist.
这只会尝试处理存在（来自ls）且格式正确的文件，而不是该范围内的每个日期/小时。如果您有稀疏文件（包括在范围的开头和结尾），这将更有效，因为它不需要检查文件是否存在。
By the way, this is the command that created the test files, if you're interested:
顺便说一下，如果您有兴趣，这是创建测试文件的命令：
pax> for dt in 20091026 20091027 20091028 ; do
for tm in 00 01 02 ... you get the idea ... 21 22 23 ; do
touch $dt.$tm.$RANDOM
done
done
Please don't type that in verbatim and then complain that it created files like:
请不要逐字输入，然后抱怨它创建了如下文件：
20091026.you.12345
20091028.idea.77
I only trimmed down the line so it fits in the code width.:-)
我只修剪了这条线，使其适合代码宽度。:-)
回答by Puppe
How about this:
这个怎么样：
#!/bin/bash
date1=
date2=
#verify dates
if ! date -d "$date1" 2>&1 > /dev/null ;
then echo "first date is invalid" ; exit 1
fi
if ! date -d "$date2" 2>&1 > /dev/null ;
then echo "second date is invalid" ; exit 1
fi
#set current and end date
current=$(date -d "$date1")
end=$(date -d "$date2 +1 hours")
#loop over all dates
while [ "$end" != "$current" ]
do
file=$(date -d "$current" +%Y%m%d.%H)
cat $file."filename" | more
current=$(date -d "$current +1 hours")
done
回答by P Shved
One possible solution: convert dates into standard Unix representation of "Seconds passed since the epoch" and loop, increasing this number by 3600 (number of seconds in an hour) each iteration. Example:
一种可能的解决方案：将日期转换为“自纪元以来经过的秒数”的标准 Unix 表示形式并循环，每次迭代将此数字增加 3600（一小时内的秒数）。例子：
#!/bin/bash
# Parse your input to date and hour first, so you get:
date_from=20090911
hour_from=10
date_to=20091026
hour_to=01
i=`date --date="$date_from $hour_from:00:00" +%s`
j=`date --date="$date_to $hour_to:00:00" +%s`
while [[ $i < $j ]]; do
date -d "1970-01-01 $i sec" "+%Y%m%d.%H"
i=$[ $i + 3600 ]
done
