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python备份文件的脚本
来源: 互联网 发布时间:2014-09-04
本文导语: 实际效果:假设给定目录"/media/data/programmer/project/python" ,备份路径"/home/diegoyun/backup/“ , 则会将python目录下的文件按照全路经备份到备份路径下,形如: /home/diegoyun/backup/yyyymmddHHMMSS/python/xxx/yyy/zzz..... 代码如下:import os import s...
实际效果:假设给定目录"/media/data/programmer/project/python" ,备份路径"/home/diegoyun/backup/“ , 则会将python目录下的文件按照全路经备份到备份路径下,形如:
/home/diegoyun/backup/yyyymmddHHMMSS/python/xxx/yyy/zzz.....
代码如下:
import os
import shutil
import datetime
def mainLogic():
#add dirs you want to copy
backdir="I:\backup"
copydirs=[]
copydirs.append("D:\programmer")
copydirs.append("D:\diegoyun")
print "Copying files ==================="
start=datetime.datetime.now()
#gen a data folder for backup
backdir=os.path.join(backdir,start.strftime("%Y-%m-%d"))
#print "backdir is:"+backdir
kc=0
for d in copydirs:
kc=kc+copyFiles(d,backdir)
end=datetime.datetime.now()
print "Finished! ==================="
print "Total files : " + str(kc)
print "Elapsed time : " + str((end-start).seconds)+" seconds"
def copyFiles(copydir,backdir):
prefix=getPathPrefix(copydir)
#print "prefix is:"+prefix
i=0
for dirpath,dirnames,filenames in os.walk(copydir):
for name in filenames:
oldpath=os.path.join(dirpath,name)
newpath=omitPrefix(dirpath,prefix)
print "backdir is:"+backdir
newpath=os.path.join(backdir,newpath)
print "newpath is:"+newpath
if os.path.exists(newpath)!=True:
os.makedirs(newpath)
newpath=os.path.join(newpath,name)
print "From:"+oldpath+" to:"+newpath
shutil.copyfile(oldpath,newpath)
i=i+1
return i
def getPathPrefix(fullpath):
#Giving /media/data/programmer/project/ , get the prefix
#/media/data/programmer/
l=fullpath.split(os.path.sep)
#print str(l[-1]=="")
if l[-1]=="":
tmp=l[-2]
else:
tmp=l[-1]
return fullpath[0:len(fullpath)-len(tmp)-1]
def omitPrefix(fullpath,prefix):
#Giving /media/data/programmer/project/python/tutotial/file/test.py ,
#and prefix is Giving /media/data/programmer/project/,
#return path as python/tutotial/file/test.py
return fullpath[len(prefix)+1:]
mainLogic()