使用bash或python删除非字母数字字符

Question

I have much files like these(please see the screenshot):

30.230201521829.jpg         
Mens-Sunglasses_L.180022111040.jpg   
progressive-sunglasses.180041285287.jpg
Atmosphere.222314222509.jpg  
Womens-Sunglasses-L.180023271958.jpg  
DAILY ESSENTIALS.211919012115.jpg
aviator-l.Sunglasses.240202216759.jpg 
aviator-l.Sunglasses.women.240202218530.jpg 

I want to raname them to the following:

230201521829.jpg          
180022111040.jpg
180041285287.jpg
222314222509.jpg
172254027299.jpg
211919012115.jpg
240202216759.jpg 
240202218530.jpg

230201521829 is a timestamp ,180022111040 is a timestamp,180041285287 is a timestamp, etc.
Ensure that the final file name looks like "timestamp.jpg".

But I am not able to write the script more.
Sed(Bash) command or Python can be used to do it?
Could you give me a example? Thanks.

Answer 1

Using command substitution for renaming the file. Following code will loop to the current directory's (unless path is modified) jpg files.

Awk is used to filter out the penultimate and last column of file name which are separated by "." .

for file in *.jpg
  do

    mv "$file" $(echo "$file" |awk -F'.' '{print $(NF-1)"." $NF}')

done

Answer 2

I use python

examp.

import os
import sys
import glob

pth = "C:\Users\Test"
dir_show = os.listdir(pth)

for list_file in dir_show:
    if list_file.endswith(".JPG"):
        (shrname, exts) = os.path.splitext(list_file)
        path = os.path.join(pth, list_file)
        newname=os.path.join(pth,shrname[shrname.find(".")+1:len(shrname)]+".JPG")
        os.rename(path,newname)

Answer 3

Using perl rename one-liner:

$ touch 30.230201521829.jpg Mens-Sunglasses_L.180022111040.jpg progressive-sunglasses.180041285287.jpg Atmosphere.222314222509.jpg Womens-Sunglasses-L.180023271958.jpg Womens-Eyeglasses-R.172254027299.jpg

$ ls -1
30.230201521829.jpg
Atmosphere.222314222509.jpg
Mens-Sunglasses_L.180022111040.jpg
progressive-sunglasses.180041285287.jpg
Womens-Eyeglasses-R.172254027299.jpg
Womens-Sunglasses-L.180023271958.jpg

$ prename -v 's/^[^.]*\.//' *.*.jpg
30.230201521829.jpg renamed as 230201521829.jpg
Atmosphere.222314222509.jpg renamed as 222314222509.jpg
Mens-Sunglasses_L.180022111040.jpg renamed as 180022111040.jpg
progressive-sunglasses.180041285287.jpg renamed as 180041285287.jpg
Womens-Eyeglasses-R.172254027299.jpg renamed as 172254027299.jpg
Womens-Sunglasses-L.180023271958.jpg renamed as 180023271958.jpg

Answer 4

You can use parameter expansion to strip off the extension, then remove all but the last . -delimited field from the remaining name. After than, you can reapply the extension.

for f in *; do
    ext=${f##*.}
    base=${f%.$ext}
    mv -- "$f" "${base##*.}.$ext"
done

The first line sets ext to the string following the last . . The second line sets base to the string that precedes the last . (by removing the last . and whatever $ext was set to). The third line constructs a new file name by first removing everything up to, and including, the final . in base , then reapplying the extension to the result.

Answer 5

#!/bin/bash/
echo "test: "
echo "" > 30.230201521829.jpg         
echo "" > Mens-Sunglasses_L.180022111040.jpg   
echo "" > progressive-sunglasses.180041285287.jpg
echo "" > Atmosphere.222314222509.jpg  
echo "" > Womens-Sunglasses-L.180023271958.jpg  
echo "" > DAILY\ ESSENTIALS.211919012115.jpg
echo "" > aviator-l.Sunglasses.240202216759.jpg 
echo "" > aviator-l.Sunglasses.women.240202218530.jpg 
echo "before: "
ls -ltr
for f in *.jpg; do
       renamed=${f: -16}
       mv "${f}" "${renamed}"
done