我有以下格式的文件:
$ ls CombinedReports_LLL-*'('*.csv
CombinedReports_LLL-20140211144020(Untitled_1).csv
CombinedReports_LLL-20140211144020(Untitled_11).csv
CombinedReports_LLL-20140211144020(Untitled_110).csv
CombinedReports_LLL-20140211144020(Untitled_111).csv
CombinedReports_LLL-20140211144020(Untitled_12).csv
CombinedReports_LLL-20140211144020(Untitled_13).csv
CombinedReports_LLL-20140211144020(Untitled_14).csv
CombinedReports_LLL-20140211144020(Untitled_15).csv
CombinedReports_LLL-20140211144020(Untitled_16).csv
CombinedReports_LLL-20140211144020(Untitled_17).csv
CombinedReports_LLL-20140211144020(Untitled_18).csv
CombinedReports_LLL-20140211144020(Untitled_19).csv
我想删除这部分:
20140211144020
(这是运行报告的时间戳,因此会有所不同)
最后是这样的:
CombinedReports_LLL-(Untitled_1).csv
CombinedReports_LLL-(Untitled_11).csv
CombinedReports_LLL-(Untitled_110).csv
CombinedReports_LLL-(Untitled_111).csv
CombinedReports_LLL-(Untitled_12).csv
CombinedReports_LLL-(Untitled_13).csv
CombinedReports_LLL-(Untitled_14).csv
CombinedReports_LLL-(Untitled_15).csv
CombinedReports_LLL-(Untitled_16).csv
CombinedReports_LLL-(Untitled_17).csv
CombinedReports_LLL-(Untitled_18).csv
CombinedReports_LLL-(Untitled_19).csv
我只是在想 mv命令,可能是这样的:
$ ls CombinedReports_LLL-*'('*.csv
但也许是 sed命令或其他会更好
最佳答案
rename
是 perl
包的一部分。它根据 perl 风格的正则表达式重命名文件。要从文件名中删除日期:
rename 's/[0-9]{14}//' CombinedReports_LLL-*.csv
如果rename
不可用,可以使用sed
+shell
:
for fname in Combined*.csv ; do mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" ; done
上面的循环遍历了你的每个文件。对于每个文件,它执行一个 mv
命令:mv "$fname""$(echo "$fname"| sed -r 's/[0-9]{14}//')"
其中,在这种情况下,sed
能够使用与上面的 rename
命令相同的正则表达式。 s/[0-9]{14}//
告诉 sed
查找连续的 14 位数字并将它们替换为空字符串。
关于bash 删除文件名的一部分,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/21691864/