删除两个字符之间的文件名bash

时间:2022-03-14 17:14:48

I have a group of files within a directory which look like:

我在目录中有一组文件,如下所示:

Filename - XXXXXXXXXXXXXXXXXXX.mp3

Where there are 19 X's. I need a bash script or similar to loop and remove the X's for all of the files within my directory.

哪里有19个X.我需要一个bash脚本或类似的循环并删除我的目录中的所有文件的X.

I have looked all over this site and others and have seen example using the sed command but I personally haven't been be to adapt it to my specific needs. I apologise for my lack of information, I am fairly new to this language and don't know where to start.

我已经浏览了这个网站和其他人,并看到了使用sed命令的示例,但我个人并不是要根据我的具体需求进行调整。我为我缺乏信息道歉,我对这种语言很新,不知道从哪里开始。

3 个解决方案

#1


0  

This is a python script that should rename all your files and remove the x's.

这是一个python脚本,应重命名所有文件并删除x。

(You should read all the lines starting with # except the 1st one)

(你应该阅读以#开头的所有行,除了第一行)

#!/usr/bin/env python
import os
for i in os.listdir("."):
    #You can change the number to whatever number of X's you want
    x = "X"*19
    #If you don't want to get rid of the ' - ', then put a '#' (without quotes) in front of the next line
    x = " - "+x
    new = i.replace(x, "")
    print(os.rename(i,new))

Save this file as XXX.py then give it the right permissions by doing:

将此文件另存为XXX.py,然后通过执行以下操作为其授予权限:

sudo chmod +x XXX.py

then run it using:

然后运行它:

./XXX.py

or if your user doesn't have the right permissions:

或者如果您的用户没有正确的权限:

sudo ./XXX.py

If the program returns a error output then:

如果程序返回错误输出,则:

Try with sudo or try:

尝试使用sudo或尝试:

sudo python XXX.py

with or without sudo.

有或没有sudo。

#2


0  

You don't really need sed for this. Just use a parameter expansion.

你真的不需要sed。只需使用参数扩展。

for file in *XXXXXXXXXXXXXXXXXXX.mp3; do
    mv "$file" "${file//XXXXXXXXXXXXXXXXXXX}"
done

If typing the Xs really bugs you…

如果输入Xs真的很烦你......

printf -v Xs 'X%.0s' {1..19}
for file in *$Xs.mp3; do
    mv "$file" "${file//$Xs}"
done

#3


0  

The Python answer might be preferable as it is treacherous to copy paste "bash" one-liners from the web, but here we go:

Python的答案可能更好,因为从网上复制粘贴“bash”单行是危险的,但是我们走了:

The following for loop will:

以下for循环将:

  1. list all mp3 files and select only those with 19 (bug alert: or more Xes before the extension)
  2. 列出所有mp3文件并仅选择那些带有19的文件(错误警报:或扩展前的更多Xes)

  3. echo each file to the screen, but removing 19 consecutive Xes first
  4. 将每个文件回显到屏幕,但首先删除19个连续的Xes

Test:

for FILE in *XXXXXXXXXXXXXXXXXXX.mp3; do echo $(echo ${FILE} | sed -E 's/X{19}//'); done

对于* XXXXXXXXXXXXXXXXXXX.mp3中的文件; do echo $(echo $ {FILE} | sed -E's / X {19} //'); DONE

This baby changes the name of the files instead of "echo":

这个宝贝改变了文件的名称而不是“echo”:

Change names:

for FILE in *XXXXXXXXXXXXXXXXXXX.mp3; do mv ${FILE} $(echo ${FILE} | sed -E 's/X{19}//'); done

对于* XXXXXXXXXXXXXXXXXXX.mp3中的文件;做mv $ {FILE} $(echo $ {FILE} | sed -E's / X {19} //'); DONE

Use with caution...

谨慎使用......

#1


0  

This is a python script that should rename all your files and remove the x's.

这是一个python脚本,应重命名所有文件并删除x。

(You should read all the lines starting with # except the 1st one)

(你应该阅读以#开头的所有行,除了第一行)

#!/usr/bin/env python
import os
for i in os.listdir("."):
    #You can change the number to whatever number of X's you want
    x = "X"*19
    #If you don't want to get rid of the ' - ', then put a '#' (without quotes) in front of the next line
    x = " - "+x
    new = i.replace(x, "")
    print(os.rename(i,new))

Save this file as XXX.py then give it the right permissions by doing:

将此文件另存为XXX.py,然后通过执行以下操作为其授予权限:

sudo chmod +x XXX.py

then run it using:

然后运行它:

./XXX.py

or if your user doesn't have the right permissions:

或者如果您的用户没有正确的权限:

sudo ./XXX.py

If the program returns a error output then:

如果程序返回错误输出,则:

Try with sudo or try:

尝试使用sudo或尝试:

sudo python XXX.py

with or without sudo.

有或没有sudo。

#2


0  

You don't really need sed for this. Just use a parameter expansion.

你真的不需要sed。只需使用参数扩展。

for file in *XXXXXXXXXXXXXXXXXXX.mp3; do
    mv "$file" "${file//XXXXXXXXXXXXXXXXXXX}"
done

If typing the Xs really bugs you…

如果输入Xs真的很烦你......

printf -v Xs 'X%.0s' {1..19}
for file in *$Xs.mp3; do
    mv "$file" "${file//$Xs}"
done

#3


0  

The Python answer might be preferable as it is treacherous to copy paste "bash" one-liners from the web, but here we go:

Python的答案可能更好,因为从网上复制粘贴“bash”单行是危险的,但是我们走了:

The following for loop will:

以下for循环将:

  1. list all mp3 files and select only those with 19 (bug alert: or more Xes before the extension)
  2. 列出所有mp3文件并仅选择那些带有19的文件(错误警报:或扩展前的更多Xes)

  3. echo each file to the screen, but removing 19 consecutive Xes first
  4. 将每个文件回显到屏幕,但首先删除19个连续的Xes

Test:

for FILE in *XXXXXXXXXXXXXXXXXXX.mp3; do echo $(echo ${FILE} | sed -E 's/X{19}//'); done

对于* XXXXXXXXXXXXXXXXXXX.mp3中的文件; do echo $(echo $ {FILE} | sed -E's / X {19} //'); DONE

This baby changes the name of the files instead of "echo":

这个宝贝改变了文件的名称而不是“echo”:

Change names:

for FILE in *XXXXXXXXXXXXXXXXXXX.mp3; do mv ${FILE} $(echo ${FILE} | sed -E 's/X{19}//'); done

对于* XXXXXXXXXXXXXXXXXXX.mp3中的文件;做mv $ {FILE} $(echo $ {FILE} | sed -E's / X {19} //'); DONE

Use with caution...

谨慎使用......