আমি এমন একটি সমাধান খুঁজছিলাম যা কোনও নতুন নির্ভরতা যুক্ত করবে না। নিম্নলিখিত (পাইথন 3) সমাধানটি দেখুন:
class VersionManager:
@staticmethod
def compare_version_tuples(
major_a, minor_a, bugfix_a,
major_b, minor_b, bugfix_b,
):
"""
Compare two versions a and b, each consisting of 3 integers
(compare these as tuples)
version_a: major_a, minor_a, bugfix_a
version_b: major_b, minor_b, bugfix_b
:param major_a: first part of a
:param minor_a: second part of a
:param bugfix_a: third part of a
:param major_b: first part of b
:param minor_b: second part of b
:param bugfix_b: third part of b
:return: 1 if a > b
0 if a == b
-1 if a < b
"""
tuple_a = major_a, minor_a, bugfix_a
tuple_b = major_b, minor_b, bugfix_b
if tuple_a > tuple_b:
return 1
if tuple_b > tuple_a:
return -1
return 0
@staticmethod
def compare_version_integers(
major_a, minor_a, bugfix_a,
major_b, minor_b, bugfix_b,
):
"""
Compare two versions a and b, each consisting of 3 integers
(compare these as integers)
version_a: major_a, minor_a, bugfix_a
version_b: major_b, minor_b, bugfix_b
:param major_a: first part of a
:param minor_a: second part of a
:param bugfix_a: third part of a
:param major_b: first part of b
:param minor_b: second part of b
:param bugfix_b: third part of b
:return: 1 if a > b
0 if a == b
-1 if a < b
"""
# --
if major_a > major_b:
return 1
if major_b > major_a:
return -1
# --
if minor_a > minor_b:
return 1
if minor_b > minor_a:
return -1
# --
if bugfix_a > bugfix_b:
return 1
if bugfix_b > bugfix_a:
return -1
# --
return 0
@staticmethod
def test_compare_versions():
functions = [
(VersionManager.compare_version_tuples, "VersionManager.compare_version_tuples"),
(VersionManager.compare_version_integers, "VersionManager.compare_version_integers"),
]
data = [
# expected result, version a, version b
(1, 1, 0, 0, 0, 0, 1),
(1, 1, 5, 5, 0, 5, 5),
(1, 1, 0, 5, 0, 0, 5),
(1, 0, 2, 0, 0, 1, 1),
(1, 2, 0, 0, 1, 1, 0),
(0, 0, 0, 0, 0, 0, 0),
(0, -1, -1, -1, -1, -1, -1), # works even with negative version numbers :)
(0, 2, 2, 2, 2, 2, 2),
(-1, 5, 5, 0, 6, 5, 0),
(-1, 5, 5, 0, 5, 9, 0),
(-1, 5, 5, 5, 5, 5, 6),
(-1, 2, 5, 7, 2, 5, 8),
]
count = len(data)
index = 1
for expected_result, major_a, minor_a, bugfix_a, major_b, minor_b, bugfix_b in data:
for function_callback, function_name in functions:
actual_result = function_callback(
major_a=major_a, minor_a=minor_a, bugfix_a=bugfix_a,
major_b=major_b, minor_b=minor_b, bugfix_b=bugfix_b,
)
outcome = expected_result == actual_result
message = "{}/{}: {}: {}: a={}.{}.{} b={}.{}.{} expected={} actual={}".format(
index, count,
"ok" if outcome is True else "fail",
function_name,
major_a, minor_a, bugfix_a,
major_b, minor_b, bugfix_b,
expected_result, actual_result
)
print(message)
assert outcome is True
index += 1
# test passed!
if __name__ == '__main__':
VersionManager.test_compare_versions()
সম্পাদনা: টুপল তুলনার সাথে বৈকল্পিক যুক্ত হয়েছে। অবশ্যই টুপল তুলনা সহ ভেরিয়েন্টটি ভাল, তবে আমি পূর্ণসংখ্যার তুলনা সহ বৈকল্পিকটি খুঁজছিলাম