'Zybooks 13.3.1 Challenge Activity

I am struggling on completing this zybook challenge activity and I don't know what else

Complete the if-else statement to print 'LOL means laughing out loud' if user_tweet contains 'LOL'.

Sample output with input: 'I was LOL during the whole movie!' LOL means laughing out loud.

python

Solution 1:^[1]

What this is asking you to do is print "LOL means laughing out loud" if LOL is in the tweet (if 'LOL' in user_tweet:).

If the tweet does not have LOL (else:), print 'No abbreviation.'

This gives you the if-else statement:

if 'LOL' in user_tweet:
    print('LOL means laughing out loud.')
else:
   print('No abbreviation.')

Solution 2:^[2]

enter code hereuser_tweet = input() if(user_tweet.find('LOL') != -1): print('LOL means laughing out loud.') else: print('No abbreviation.')

Using the find function works as well. The find function is searching user_tweet for string placement index != (not equal) to -1. The find function returns the index of the first occurrence of item x in the string, else returns -1.

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Source: Stack Overflow

Solution	Source
Solution 1	Tomerikoo
Solution 2	hawkisberg

'Zybooks 13.3.1 Challenge Activity

Solution 1:[1]

Solution 2:[2]

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Related Questions

Solution 1:^[1]

Solution 2:^[2]