我有以下文件列表:
lab-1:/tmp/tmp.NMFaDO# ls
webserver.123.net.1970-01-01.csv webserver.123.net.2016-10-01.csv webserver.123.net.2017-10-01.csv
webserver.123.net.2015-11-01.csv webserver.123.net.2016-11-01.csv webserver.123.net.2017-11-01.csv
webserver.123.net.2015-12-01.csv webserver.123.net.2016-12-01.csv webserver.123.net.2017-12-01.csv
webserver.123.net.2016-01-01.csv webserver.123.net.2017-01-01.csv webserver.123.net.2018-01-01.csv
webserver.123.net.2016-02-01.csv webserver.123.net.2017-02-01.csv webserver.123.net.2018-02-01.csv
webserver.123.net.2016-03-01.csv webserver.123.net.2017-03-01.csv webserver.123.net.current.csv
webserver.123.net.2016-04-01.csv webserver.123.net.2017-04-01.csv webserver.123.net.db.done
webserver.123.net.2016-05-01.csv webserver.123.net.2017-05-01.csv webserver.123.net.records_to_delete.txt
webserver.123.net.2016-06-01.csv webserver.123.net.2017-06-01.csv sip2.ci2.lab.jwm2.net.2018-02-01.csv
webserver.123.net.2016-07-01.csv webserver.123.net.2017-07-01.csv sip2.ci2.lab.jwm2.net.db.done
webserver.123.net.2016-08-01.csv webserver.123.net.2017-08-01.csv sip2.ci2.lab.jwm2.net.records_to_delete.txt
webserver.123.net.2016-09-01.csv webserver.123.net.2017-09-01.csv
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在 bash 脚本中,我需要遍历此列表并仅提取文件名的日期部分。我设法弄清楚如何从
包含使用此命令的文件的文件夹中执行此操作:
lab-1:/tmp/tmp.NMFaDO# ls webserver.123.net*.csv | sed 's/webserver.123.net.\(.*\)\.csv/\1/'
1970-01-01
2015-11-01
2015-12-01
2016-01-01
2016-02-01
2016-03-01
2016-04-01
2016-05-01
2016-06-01
2016-07-01
2016-08-01
2016-09-01
2016-10-01
2016-11-01
2016-12-01
2017-01-01
2017-02-01
2017-03-01
2017-04-01
2017-05-01
2017-06-01
2017-07-01
2017-08-01
2017-09-01
2017-10-01
2017-11-01
2017-12-01
2018-01-01
2018-02-01
current
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但我似乎无法让逻辑在我的 bash 脚本中工作:
dir=/tmp/tmp.NMFaDO
server=webserver.123.net
for csvfile in $dir/$server.*.csv
do
echo $csvfile | sed 's/$dir\/$server.\(.*\)\.csv/\1/'
done
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它目前正在打印名称的完整路径,如下所示:
/tmp/tmp.NMFaDO/webserver.123.net.1970-01-01.csv
/tmp/tmp.NMFaDO/webserver.123.net.2015-11-01.csv
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等等。
有人可以告诉我我的 bash 脚本中缺少什么吗?谢谢。
使用这样的文件名格式,您可以删除扩展名,然后删除最后一个点之前的所有内容。
带参数扩展:
for f in ./webserver.123.*; do
d=${f%.csv} # remove trailing .csv
d=${d##*.} # remove all up to (now) last .
echo "filename: $f date: $d"
done
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或使用 sed:
for f in ./webserver.123.*; do
d=$(echo "$f" | sed -e 's/\.csv$//' -e 's/.*\.//');
echo "filename: $f date: $d";
done
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