Linux下 rename修改后缀文件扩展名 cp复制多个文件

rename .txt .csv *

find . -type f -exec mv {} {}.xml ';'

cp /home/usr/dir/{file1,file2,file3,file4} /home/usr/destination/

echo 'a1 a2 a3' | xargs -n 1 cp a

cp /home/usr/dir/file{1..4} ./

import os,sys,shutil
### copies a list of files from source. handles duplicates.
def rename(file_name, dst, num=1):
    #splits file name to add number distinction
    (file_prefix, exstension) = os.path.splitext(file_name)
    renamed = "%s(%d)%s" % (file_prefix,num,exstension)
 
    #checks if renamed file exists. Renames file if it does exist.
    if os.path.exists(dst + renamed):
        return rename(file_name, dst, num + 1)
    else:
        return renamed
 
def copy_files(src,dst,file_list):
    for files in file_list:
        src_file_path = src + files
        dst_file_path = dst + files
        if os.path.exists(dst_file_path):
            new_file_name =  rename(files, dst)
            dst_file_path = dst + new_file_name
 
        print "Copying: " + dst_file_path
        try:
            # 复制操作主要就是这句
            shutil.copyfile(src_file_path,dst_file_path)
        except IOError:
            print src_file_path + " does not exist"
            raw_input("Please, press enter to continue.")
 
def read_file(file_name):
    f = open(file_name)
    #reads each line of file (f), strips out extra whitespace and 
    #returns list with each line of the file being an element of the list
    content = [x.strip() for x in f.readlines()]
    f.close()
    return content
 
src = sys.argv[1]
dst = sys.argv[2]
file_with_list = sys.argv[3]
 
copy_files(src,dst,read_file(file_with_list))