Better way to solve Welp Step 9?

gr8 work just figuring out how could u think all this :heart_eyes:

Nice one. I got the same answer. Here is my query:

WITH combined AS (
SELECT reviews.username, reviews.rating, places.name, places.average_rating
FROM places
JOIN reviews
ON places.id = reviews.place_id
)
SELECT username, COUNT(*) AS below_ratings
FROM combined
WHERE rating < average_rating
GROUP BY username
ORDER BY below_ratings DESC
LIMIT 1;

4 Likes

I did the same. The only thing I think we should consider that the column average_rating is not the actual average rating for all places, but is the average rating for a single restaurant. I think with a little tweak the statement should be true.

SELECT reviews.username, COUNT (*) FROM reviews

JOIN places

ON reviews.place_id = places.id

WHERE reviews.rating < (SELECT AVG (average_rating) FROM places)

GROUP BY 1

ORDER BY 2 DESC

LIMIT 1;

1 Like

This is actually why I think it’s the best column to use and it isn’t necessary to average the average. The reason is because I feel a customer is only being a difficult reviewer if their review is below that particular place’s average. If “Bob’s Expired Bistro” opens up and gets an average rating of a 2, it doesn’t mean that someone rating it a 2 is being harsh - the person is accurately rating that particular restaurant. However, if someone rates a restaurant a 1 when that restaurant’s average is a 2, then they are being more difficult than the average patron of that establishment.

That’s my opinion & interpretation of the question, at least. They don’t provide us a solution or a desired output, so it’s open to our own ideas on it.

5 Likes

That’s a valid point. “Average rating for places” is a little vague and is a question of interpretation. Thanks.

2 Likes

Hey,

So I was stuck as well, but still got pinkdeb with 7 bad reviews (first block in the file below).

Then I realized that yes they may be the one with the higher number of bad reviews but we don’t know for sure as we didn’t compare their % of bad reviews on the total of reviews.

I came up with this (second block).

If we remove the ones with only 1 review, our worst user is actually youngNOTold with 85% of bad reviews.

I put LIMIT 15 instead of 1 to validate my point, don’t get me wrong!

hey guys here is what i did, let me know if makes sense :slight_smile:

SELECT R.username, COUNT(*)

FROM reviews AS R

INNER JOIN places AS P

ON R.place_id = P.id

WHERE R.rating < P.average_rating

GROUP BY R.username

ORDER BY COUNT(*) DESC;

Here’s what I wrote. Where did I go wrong? Thanks in advance.


WITH Average_Rating AS (

SELECT AVG(average_rating) AS Avg_Rating

FROM places

),

Bad_Reviews AS (

SELECT *

FROM reviews

CROSS JOIN Average_Rating

WHERE rating < Average_Rating 

),

Bad_Reviews_by_User AS (

SELECT username,

  COUNT(id) AS '#_of_Bad_Reviews'

FROM Bad_Reviews

GROUP BY 1

ORDER BY 2 DESC

LIMT 1;

)

Maybe it’s just that typo :slight_smile:

1 Like

Even after I corrected the spelling, I still don’t see anything.

I checked your query again.
You created all your temporary tales (Average_Rating, Bad_Reviews and Bad_Reviews_By_User) using WITH, but you never actually selected something from any of them.

1 Like

It’s great seeing all different ways people have solved this step! I didn’t limit to 1 and select only the username because I wanted more info to confirm I was getting the right query results.

SELECT reviews.username, COUNT(reviews.username) AS ‘review_count’, reviews.rating, places.average_rating
FROM places
JOIN reviews
ON places.id = reviews.place_id
GROUP BY reviews.username
HAVING reviews.rating < places.average_rating
ORDER BY COUNT(reviews.username) DESC;

I got the same answer with a different approach. Any feedback would be appreciated!

SELECT reviews.rating, places.average_rating, reviews.username, COUNT()
FROM reviews
JOIN places
ON reviews.place_id = places.id
WHERE reviews.rating < places.average_rating
GROUP BY username
ORDER BY COUNT(
) DESC
LIMIT 1;

Thanks, all!

Hi @raquelaurelia1999 ,
I was on the look for verifying the correctness of my solution.
Below is my approach.
I thought about looking only for reviewers who left negative feedback.
Besides that my solution is very similar to yours.
Let’s call this a win-win :slight_smile:

SELECT reviews.username, reviews.rating, reviews.note, places.average_rating
, COUNT(reviews.username)
FROM reviews
INNER JOIN places
  ON reviews.place_id = places.id
WHERE  reviews.rating < 2.5
GROUP BY reviews.username
-- ORDER BY reviews.username;
ORDER BY COUNT(reviews.username) DESC;

I think I read the question a little differently. I read it as the person that left the most below average reviews to ONE place.


SELECT places.name, reviews.username, COUNT(reviews.rating)
FROM reviews
JOIN places
ON reviews.place_id = places.id
WHERE reviews.rating < places.average_rating
GROUP BY reviews.username, places.name
ORDER BY COUNT(reviews.rating) DESC
LIMIT 1;

SELECT username, COUNT() AS ‘Number of Reviews’, ROUND(AVG(rating),1) AS ‘User Average Rating’, average_rating AS ‘Place Average Rating’

FROM places

JOIN reviews

ON places.id = reviews.place_id

WHERE places.average_rating > reviews.rating

GROUP BY 1

ORDER BY 2 DESC

LIMIT 1;

I did this and gave me the correct solution, I am a newbie so if anybody finds something wrong let me know :slight_smile:

SELECT reviews.username, COUNT ()

FROM places

INNER JOIN reviews

ON places.id = reviews.place_id

WHERE reviews.rating < places.average_rating

GROUP BY reviews.username

ORDER BY COUNT () DESC;

Hello, @blog7064965569 !
I finished this task recently and came here for any details and other opinons. And when I saw your solution I came up with a question, if you don’t mind.
As I understood the “HAVING” clause filters the groups of data. So your “GROUP” clause collects rows from reviews table with the same username into the groups. And each row within the group is a single review and has a different rating value.
If so, which rating value of the rows in the group are you comparing with average_rating ? Because it seems to me that when using “HAVING” clause you can compare characteristics that are the same for the whole group. And rating value doesn’t seem like that.

I hope I’m understandable and I would be glad to have any feedback! :slight_smile:

SELECT 
  reviews.username, COUNT(*)
FROM reviews
JOIN places
  ON reviews.place_id = places.id
  WHERE reviews.rating < places.average_rating
  GROUP BY 1
  ORDER BY 2 DESC
  LIMIT 1;

This is what I ended up with. Any thoughts?