so i want to avoid overwrite the file name that existed. but i don't know how to combine the code with mycode. please help me
here's my code for write file:
def filepass(f):
print(f)
with open ('media/pass/'+'filepass.txt', 'a') as fo:
fo.write(f)
fo.close()
return fo
and here's the code to create number in name filepass:
def build_filename(name, num=0):
root, ext = os.path.splitext(name)
print(root)
return '%s%d%s' % (root, num, ext) if num else name
def find_next_filename(name, max_tries=20):
if not os.path.exists(name): return name
else:
for i in range(max_tries):
test_name = build_filename(name, i+1)
if not os.path.exists(test_name): return test_name
return None
all i want is to create filename: filepass.txt, filepass1.txt, filepass2.txt
Something like this?
def filepass(f):
print(f)
filename = find_next_filename('media/pass/filepass.txt')
with open (filename, 'a') as fo:
fo.write(f)
# you don't need to close when you use "with open";
# but then it doesn't make sense to return a closed file handle
# maybe let's return the filename instead
return filename
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