os.path.split，改变文件名与妥协的Path [英] os.path.split, changing file name with out compromising the Path

问题描述

我已经遵循 Python获取文件名并更改&保存在变量中。哪些工作正常，并根据需要更改文件名。

但现在我面临文件被保存的路径问题。文件被保存在 media / ok_abc.txt 中，而应该是 media / documents / ok_abc.txt

$ b $例如

docfile = /media/documents/abc.csv
在应用下面说明

  filename = os.path.splitext（docfile.name）[0] 
 newfilename ='ok_％s.txt'％filename 
  pre> 
 
 
可以更改文件名，但路径随着 /media/ok_abc.txt 而减少，应该是 / media/documents/abc.txt 
 
 
 
我可以如何在路径上修改文件名来妥协。
解决方案
从完整的文件路径中提取目录，然后将其添加回来。
  path，filename = os.path.split（docfile）
 filename = os.path.splitext（filename）[0] 
 newfilename ='ok_％s.txt'％filename 
 newpath = os.path.join（path，newfilename）
  
 
i have followed Python get file name and change & save it in variable. which work fine and change the file name as required. 

but now i am facing problem with path where the file is getting saved. as the file is getting saved in "media/ok_abc.txt" whereas it should be media/documents/ok_abc.txt

e.g. 

docfile = /media/documents/abc.csv
after applaying below instruction
filename = os.path.splitext(docfile.name)[0]
newfilename = 'ok_%s.txt' % filename
am able to change the file name but the path is getting reduced as /media/ok_abc.txt, it should be /media/documents/abc.txt

how i can change the file name with out compromising on the Path
解决方案
Extract the directory from the full file path, and later add it back.
path, filename = os.path.split(docfile)
filename = os.path.splitext(filename)[0]
newfilename = 'ok_%s.txt' % filename
newpath = os.path.join(path, newfilename)


                        
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