bash删除文件名的一部分 [英] bash removing part of a file name

问题描述

我具有以下格式的以下文件:

I have the following files in the following format:

$ 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

我要删除此部分:
20140211144020 (这是运行报告的时间戳记，因此会有所不同)

I would like this part removed:
20140211144020 (this is the timestamp the reports were run so this will vary)

并以类似以下内容结束:

and end up with something like:

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

我只是按照的思路思考mv 命令，也许像这样:

I was thinking simply along the lines of the mv command, maybe something like this:

$ ls CombinedReports_LLL-*'('*.csv

但 sed 命令或其他命令会更好

but maybe a sed command or other would be better

推荐答案

重命名是 perl 软件包的一部分.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，请执行以下操作:

rename is part of the perl package. It renames files according to perl-style regular expressions. To remove the dates from your file names:

rename 's/[0-9]{14}//' CombinedReports_LLL-*.csv

如果没有 rename ，则可以使用 sed + shell :

If rename is not available, sed+shell can be used:

for fname in Combined*.csv ; do mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" ; done

以上内容循环遍历您的每个文件.对于每个文件，它执行 mv 命令: mv"$ fname"" $(echo"$ fname" | sed -r's/[0-9] {14}//')'其中，在这种情况下， sed 是能够使用与上述 rename 命令相同的正则表达式. s/[0-9] {14}//告诉 sed 连续查找14位数字并将其替换为空字符串.

The above loops over each of your files. For each file, it performs a mv command: mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" where, in this case, sed is able to use the same regular expression as the rename command above. s/[0-9]{14}// tells sed to look for 14 digits in a row and replace them with an empty string.

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