【python】獲取目錄下的最新資料夾/檔案
阿新 • • 發佈:2021-01-08
技術標籤:Project Managerpython AI BlockChain
def new_report(test_report): lists = os.listdir(test_report) #列出目錄的下所有檔案和資料夾儲存到lists print(list) lists.sort(key=lambda fn:os.path.getmtime(test_report + "\\" + fn))#按時間排序 file_new = os.path.join(test_report,lists[-1]) #獲取最新的檔案儲存到file_new print(file_new) return file_new if __name__=="__main__": test_report="path"#目錄地址 new_report(test_report)
df_data=dataPress(fileName) # os.remove(u'./upload/{}'.format(fileName)) def get_filelist(dir, Filelist): try: lists = os.listdir(dir) lists.sort(key=lambda fn:os.path.getmtime(dir + "\\" + fn)) if len(lists)>1: lists=[i for i in lists if (R"xlsx" in i)] for i in lists: print(i) if i !=lists[-1] or R"xlsx" not in i : os.remove(u'./upload/{}'.format(i)) file_new = lists[-1] print(file_new) if R"xlsx" in file_new: return file_new # for s in os.listdir(dir): # if R"xlsx" in s: # print(s) # Filelist.append(s) # return Filelist[0] except Exception,err: print(err) return Filelist def dataPress(fileName): print("done >>>>>>>>>>",fileName.decode("utf-8")) try: excel_reader=pd.ExcelFile(u'./upload/{}'.format(fileName)) except Exception,err: if R"xlsx" in str(err): df=pd.ExcelFile(u'./upload/{}'.format(fileName),engine='openpyxl') sheet_names = excel_reader.sheet_names lis=[] if R"dataflow" in sheet_names: df_data = excel_reader.parse(sheet_name="dataflow") df_=df_data.iloc[1:5,1:5] df_.iloc[:,2:4] = np.nan for tup in df_.itertuples(): lis.append(tup[1:]) # print(tup[1:]) print(lis) return lis # result = _common.mysql_select(sql, withcolumns=False)