python - 使用多个 if 条件时如何克服 ValueError？

Question

我正在尝试制作一个脚本来识别文件夹名称是否是项目。为此，我想使用多个 if 条件。但我对检查文件夹名称的第一个字母以及它是否为数字等结果产生的 ValueError 感到困扰。如果它是一个字符串，我想跳过该文件夹并使其检查下一个文件夹。预先感谢大家的帮助。 干杯，本

我尝试了 While 和 except ValueError: 但没有成功。

# 正确的项目名称 "YYMM_ProjectName"= "1908_Sample_Project"

projectnames = ['190511_Waldfee', 'Mountain_Shooting_Test', '1806_Coffe_Prime_Now', '180410_Fotos', '191110', '1901_Rollercoaster_Vision_Ride', 'Musicvideo_LA', '1_Project_Win', '19_Wrong_Project', '1903_di_2', '1907_DL_2', '3401_CAR_Wagon']

# 检查条件

for projectname in projectnames:
    if int(str(projectname[0])) < 3 and int(projectname[1]) > 5 and ((int(projectname[2]) * 10) + int(projectname[3])) <= 12 and str(projectname[4]) == "_" and projectname[5].isupper():
        print('Real Project')
        print('%s is a real Project' % projectname)
        # print("Skipped Folders")

ValueError:以 10 为基数的 int() 的文字无效:'E'

Answer 1

根据我对所有 ifs 的理解...您实际上使用 regex 可能会更好匹配。您正在解析每个字符，并期望每个字符都在非常有限的字符范围内。

我尚未测试此模式字符串，因此它可能不正确或需要根据您的需要进行调整。

import re

projectnames = ['1911_Waldfee', "1908_Project_Test", "1912_WinterProject", "1702_Stockfootage", "1805_Branded_Content"]

p = ''.join(["^", # Start of string being matched
             "[0-2]", # First character a number 0 through 2 (less than 3)
             "[6-9]", # Second character a number 6 through 9 (single digit greater than 5)
             "(0(?=[0-9])|1(?=[0-2]))", # (lookahead) A 0 followed only by any number 0 through 9 **OR** A 1 followed only by any number 0 through 2
             "((?<=0)[1-9]|(?<=1)[0-2])", # (lookbehind) Match 1-9 if the preceding character was a 0, match 0-2 if the preceding was a 1
             "_", # Next char is a "_"
             "[A-Z]", #Next char (only) is an upper A through Z
              ".*$" # Match anything until end of string 
    ])

for projectname in projectnames:
    if re.match(p, projectname):
        #print('Real Project')
        print('%s is a real Project' % projectname)
        # print("Skipped Folders")

编辑:==========================

您可以使用以下方法逐步测试该模式...

projectname = "2612_UPPER"
p = "^[0-2].*$" # The first character is between 0 through 2, and anything else afterwards
if re.match(p, projectname): print(projectname) 
# If you get a print, the first character match is right.
# Now do the next
p = "^[0-2][6-9].*$" # The first character is between 0 through 2, the second between 6 and 9, and anything else afterwards
if re.match(p, projectname): print(projectname) 
# If you get a print, the first and second character match is right.
# continue with the third, fourth, etc.