In my test code, I want to assert that a string ends with a number. Say the number is between [0,3):
assert_equals('/api_vod_asset/v0/assets/0', '/api_vod_asset/v0/assets/number') #valid
assert_equals('/api_vod_asset/v0/assets/1', '/api_vod_asset/v0/assets/number') #valid
assert_equals('/api_vod_asset/v0/assets/5', '/api_vod_asset/v0/assets/number') #invalid
How to use regular expression or some other technique for number ?
You would probably want to use assertRegex:
test_case.assertRegex('/api_vod_asset/v0/assets/0', '/api_vod_asset/v0/assets/[012]')
The one above works in the case of [0,3) range. If you do not want that restriction, you would likely want to have:
test_case.assertRegex('/api_vod_asset/v0/assets/0', '/api_vod_asset/v0/assets/[\d]')
All the code above works after the following lines have been added to your snippet:
import unittest as ut
test_case = ut.TestCase()
If it will always be at the same place in the string, you can use string.split
something like
def check_range(to_test, valid_range):
return int(to_test.split('/')[-1]) in range(valid_range[0], valid_range[1] + 1)
then you can do
check_range('/api_vod_asset/v0/assets/0', [0,3]) #True
check_range('/api_vod_asset/v0/assets/1', [0,3]) #True
check_range('/api_vod_asset/v0/assets/5', [0,3]) #False
First, use a regex like \d+$ to get any number (\d+) at the end of the string ($)...
>>> m = re.search(r"\d+$", s)
... then check whether you have a match and whether the number is in the required range:
>>> m is not None and 0 <= int(m.group()) < 3
Or use a range, if you prefer this notation (assuming upper bound of [0,3) is exclusive):
>>> m is not None and int(m.group()) in range(0, 3)
re support \D? Then use \D[0-3]$ :)
I want to assert that a string ends with a number
if int(myString[-1]) in [0,1,2]:
do something...
myString[-1] gives you only the last char of the string. To get 10 you need myString[-2:]. However, if you want to check 'any number of chars after the last '/' you can use myString.split('/')[-1] instead. Assuming that your strings will have the number after the last slash.>>> import re
>>> match = re.search(r'/api_vod_asset/v0/assets/(\d+)', '/api_vod_asset/v0/assets/5')
>>> match
<_sre.SRE_Match object at 0x10ff4e738>
>>> match.groups()
('5',)
>>> match.group(1)
'5'
>>> x = int(match.group(1))
>>> 0 <= x < 3
False
>>> 0 <= 2 < 3
True
Regex:
import re
reg_func = lambda x: bool(re.match(r'.*?[0123]$', x))
Python:
py_func = lambda x: x.endswith(tuple(map(str, range(4))))
I'm not sure that this is the most elegant way to do this but here is what i would do:
def is_finished_correctly(path):
# The code seems self-explaining, ask if you need more detailed explanations
return (int(path.split("/")[-1]) in range(3))
assert(is_finished_correctly('/api_vod_asset/v0/assets/0'))
If you don't necessarily want to use regex, try this:
a = "hello"
a[-1] = "o" # gives you the last character of a string
try:
int(a[-1])
print "last position is int"
except:
print "last position is no int"
int() raises a ValueError when you pass a parameter in a wrong base, which is 10 per default.
You could now do something like this:
path = '/api_vod_asset/v0/assets/0'
try:
int(path[-1])
# do all your stuff here
except:
print "wrong path"
continue # (if you are in a loop)
(If you loop over many paths, some of which might not be want you want to access)
