Challenge: The wordiest author

It might be because your total_words needs another zero, to be 1 million. If it doesn't work after that, I had a problem where I thought my query was correct but it wasn't letting me move on. So I copied my query, clicked on the Start Over button, and then pasted my query back in and it worked.

select author, sum(words) as total_words from books group by author
having total_words >=1000000;

i'm also facing the same problem.please suggest me answer.

You dont have to use the SUM(words) here. Nobody's asking to add the words or anything. Instead use TOTAL(words) and use the query as follows;

SELECT author, TOTAL(words) AS total_words FROM books GROUP BY author HAVING total_words > 1000000

Should work.

Yes your query needs a 0

what about the 2nd step?

select author, avg(words) as avg_words from books group by author having avg_words=150000;

how to choose per book?

For me the problem was that I'd edited the answer to the first problem for the second rather than adding it as a new query.  Both Queries need to be entered in order to pass this part.

Query #1

SELECT author, SUM(words) AS total_words FROM books GROUP BY author HAVING total_words > 1000000;

Query #2

SELECT author, AVG(words) AS avg_words FROM books GROUP BY author HAVING avg_words>150000;

Also make sure that you haven't done any other queries before the one that it is checking. I spent a solid 30 minutes typing in the correct query over and over and double-checked it against this thread, only to figure out that an extra query of SELECT * FROM books was throwing everything off; I just wanted to see the whole table for reference.

DrafGracula: 
select author, avg(words) as avg_words from books group by author having avg_words >=150000

Actually sometimes their code playground was not working in mobiles even though you give correct input because as i started writing leaving two lines then select my and was not passing rather if i don't give empty lines the test passed so weird but the results displayed,
By the way answers are

SELECT author,SUM(words) AS total_words FROM books GROUP BY author HAVING total_words > 1000000;
SELECT author,AVG(words) AS avg_words FROM books GROUP BY author HAVING avg_words > 150000;

Yes the following both answers work by replacing the sum with total it also gives the same answer

SELECT author, SUM(words)as total_words FROM books GROUP BY author HAVING total_words>1000000;

SELECT author, TOTAL(words)as total_words FROM books GROUP BY author HAVING total_words>1000000;

I just did this:

one query ONLY

make sure to start over with each mistake...

SELECT author, SUM(words) AS total_words FROM books GROUP BY author HAVING total_words > 1000000;

Hi. I was wondering about writing blogs of my site and found this post. I do not see the email info for Khan on this website. Do you have the details? 

select author, sum(words) As total_words from books
group by author having total_words > 1000000;

select author, AVG(words) As avg_words from books
group by author having avg_words > 150000;