python 30课习题，腌制文件 报错，求详解
'''小甲鱼源代码报错，求详解Traceback (most recent call last):
File "C:\Users\admin\Desktop\python.py", line 40, in <module>
split_file('record.txt')
File "C:\Users\admin\Desktop\python.py", line 21, in split_file
f = open(file_name)
FileNotFoundError: No such file or directory: 'record.txt'''
#下面是小甲鱼源代码
import pickle
def save_file(boy, girl, count):
file_name_boy = 'boy_' + str(count) + '.txt'
file_name_girl = 'girl_' + str(count) + '.txt'
boy_file = open(file_name_boy, 'wb') # 记得一定要加 b 吖
girl_file = open(file_name_girl, 'wb') # 记得一定要加 b 吖
pickle.dump(boy, boy_file)
pickle.dump(girl, girl_file)
boy_file.close()
girl_file.close()
def split_file(file_name):
count = 1
boy = []
girl = []
f = open(file_name)
for each_line in f:
if each_line[:6] != '======':
(role, line_spoken) = each_line.split(':', 1)
if role == '小甲鱼':
boy.append(line_spoken)
if role == '小客服':
girl.append(line_spoken)
else:
save_file(boy, girl, count)
boy = []
girl = []
count += 1
save_file(boy, girl, count)
f.close()
split_file('record.txt')
FileNotFoundError: No such file or directory: 'record.txt'''
错误告诉你了文件未找到，应该是你路径错了。
如果文件不是跟你的py文件在同一目录下的话，请输入完整路径。 f = open(file_name) 这里没有设置打开模式, Python就会使用默认的'r'模式进行打开, 而'r'模式下, 文件不存在的话就会报错.
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