bash bash删除文件名的一部分
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bash removing part of a file name
提问by HattrickNZ
I have the following files in the following format:
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
I would like this part removed: 20140211144020(this is the timestamp the reports were run so this will vary)
我希望删除这部分：（20140211144020这是运行报告的时间戳，因此会有所不同）
and end up with something like:
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
I was thinking simply along the lines of the mvcommand, maybe something like this:
我只是按照mv命令的思路思考，也许是这样的：
$ ls CombinedReports_LLL-*'('*.csv
but maybe a sedcommand or other would be better
但也许sed命令或其他命令会更好
回答by John1024
renameis part of the perlpackage. It renames files according to perl-style regular expressions. To remove the dates from your file names:
rename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
If renameis not available, sed+shellcan be used:
如果rename不可用，可以使用sed+ shell：
for fname in Combined*.csv ; do mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" ; done
The above loops over each of your files. For each file, it performs a mvcommand: mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')"where, in this case, sedis able to use the same regular expression as the renamecommand above. s/[0-9]{14}//tells sedto look for 14 digits in a row and replace them with an empty string.
以上循环遍历您的每个文件。对于每个文件，它执行一个mv命令： mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')"在这种情况下，sed可以使用与上述rename命令相同的正则表达式。 s/[0-9]{14}//告诉sed在一行中查找 14 位数字并用空字符串替换它们。
回答by Alfe
Without using an other tools like renameor sedand sticking strictly to bashalone:
不使用其他工具，如rename或sed并严格坚持bash单独使用：
for f in CombinedReports_LLL-*.csv
do
newName=${f/LLL-*\(/LLL-(}
mv -i "$f" "$newName"
done
回答by Pedro Nunes
for f in CombinedReports_LLL-* ; do
b=${f:0:20}${f:34:500}
mv "$f" "$b"
done
You can try line by line on shell:
您可以在 shell 上逐行尝试：
f="CombinedReports_LLL-20140211144020(Untitled_11).csv"
b=${f:0:20}${f:34:500}
echo $b
回答by user5318429
I'm using the advice given in the top response and have put the following line into a shell script:
我正在使用顶部响应中给出的建议，并将以下行放入 shell 脚本中：
ls *.nii | xargs rename 's/[f_]{2}//' f_0*.nii
In terminal, this line works perfectly, but in my script it will not execute and reads * as a literal part of the file name.
在终端中，这一行工作得很好，但在我的脚本中，它不会执行并将 * 作为文件名的文字部分读取。
回答by Kevin
You can use the renameutility for this. It uses syntax much like sed to change filenames. The following example (from the rename man-page) shows how to remove the trailing '.bak' extension from a list of backup files in the local directory:
您可以rename为此使用该实用程序。它使用类似于 sed 的语法来更改文件名。以下示例（来自重命名手册页）显示了如何从本地目录中的备份文件列表中删除尾随的“.bak”扩展名：
rename 's/\.bak$//' *.bak
