bash删除文件名的一部分

发布时间：2020-12-16 01:16:40 所属栏目：安全来源：网络整理

导读：我有以下格式的以下文件： $ls CombinedReports_LLL-*'('*.csvCombinedReports_LLL-20140211144020(Untitled_1).csvCombinedReports_LLL-20140211144020(Untitled_11).csvCombinedReports_LLL-20140211144020(Untitled_110).csvCombinedReports_LLL-20140211

我有以下格式的以下文件：

$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(这是报告运行的时间戳,因此会有所不同)

并最终得到类似的东西：

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

我只是按照mv命令的思路思考,可能是这样的：

$ls CombinedReports_LLL-*'('*.csv

但也许sed命令或其他更好

重命名是perl包的一部分.它根据perl样式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

如果重命名不可用,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} //’)”where,在这种情况下,sed能够使用与上面的重命名命令相同的正则表达式. s / [0-9] {14} //告诉sed连续查找14位数字并用空字符串替换它们.

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!