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粉碎 选择从某一方位起的档案
原标题:Python: Choose a file starting with a given string
  • 时间:2023-12-20 02:41:07
  •  标签:
  • python

我有一个疑问,即从一个文件夹中取出一个文件,将其移至另一个文件夹,并按日期制作一个文件夹,但我不得不对一个文件提出类似的询问,这些档案是从某个名字开始的,而Im没有找到使用再研究的适当方法:

文档名称为GM_+4位数(项目编号)++日期(YYYMMDD),例如: GM_0435_20231220

files = os.listdir(sourceDir)
for file in files:
    GM = re.search("$GM", file)
    if GM:
        
        project = file[4:7]
        date = file[0][:8]
        year = date[:4]
        month = date[-2:6]

利用原档案的一部分(日期),设法在册子中找到一个外壳电子表格,将其移至某一倍数的圆顶点,从中点取分离的立体卷宗。

问题回答

查阅所有档案的os.listdir(>>,可使用glob.glob(,其中只有与特定类型相匹配的文档:

import glob
for filename in glob.glob( GM_[0-9][0-9][0-9][0-9]_[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9] ):
    # filename is guaranteed to be a match here...

If you like, you can use string multiplication to make the pattern a bit more readable:

d =  [0-9] 
for filename in glob.glob( GM_  + d*4 +  _  + d*8):




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