hey guys here is what i did, let me know if makes sense
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;
hey guys here is what i did, let me know if makes sense
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
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.
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
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
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!
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?
Why can’t we use > WHERE reviews.rating < (SELECT AVG (average_rating) FROM places)
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1;
yizuhiProblem Solver (bronze)
selectall
Oh. So maybe it’s just because I understood the question differently. (Still, I think I can find a faster way to do what I did).
I like the way you approached it, and your query is so clean
Actually, mine was the same as yours. (Guess he really is a difficult reviewer no matter how you approach things )
Thanks for your comment, it’s so nice to get to understand the way somebody else approaches a problem
1
Reply
16 DAYS LATER
Hi,
I had the same thought as yours and took some struggle to complete this one, finally I managed to complete it in a similar way to yours. I was able to complete using an inner query, giving me a slightly
“cleaner” solution:
with bad_reviews AS (
select * from reviews
join places on
reviews.place_id = places.id
)
select username,count(*) from bad_reviews
where rating < (
select avg(average_rating)
from bad_reviews
)
group by username
order by count(*) desc
;
1
Reply
1 MONTH LATER
selectall
Good one i too did sth like the one u did.
username = @pinkdeb
review_count(bad) = 7
Reply
gr8 work just figuring out how could u think all this
Reply
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
Reply
8 DAYS LATER
selectall
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 Reply
1
Reply
selectallBeta Tester
marinsimeonoff346770
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.
1 Reply
5
Reply
selectall
That’s a valid point. “Average rating for places” is a little vague and is a question of interpretation. Thanks.
2
Reply
19 DAYS LATER
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.
1 Reply
Reply
valriecloutier078343
I put LIMIT 15 instead of 1 to validate my point, don’t get me wrong!
Reply
2 MONTHS LATER
hey guys here is what i did, let me know if makes sense
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;
Reply
29 DAYS LATER
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;
)
…
1 Reply
Reply
yizuhiProblem Solver (bronze)
gguirao13
Maybe it’s just that typo
1 Reply
1
Reply
yizuhi
Even after I corrected the spelling, I still don’t see anything.
1 Reply
Reply
yizuhiProblem Solver (bronze)
gguirao13
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
Reply
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;
2 Replies
Reply
19 DAYS LATER
raquelaurelia1999Today is the anniversary of the day I joined this community!
blog7064965569
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!
Reply
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
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;
Reply > WHERE reviews.rating < (SELECT AVG (average_rating) FROM places)
GROUP BY 1
ORDER BY 2 DESC
LIMIT 1;
yizuhiProblem Solver (bronze)
selectall
Oh. So maybe it’s just because I understood the question differently. (Still, I think I can find a faster way to do what I did).
I like the way you approached it, and your query is so clean
Actually, mine was the same as yours. (Guess he really is a difficult reviewer no matter how you approach things )
Thanks for your comment, it’s so nice to get to understand the way somebody else approaches a problem
1
Reply
16 DAYS LATER
Hi,
I had the same thought as yours and took some struggle to complete this one, finally I managed to complete it in a similar way to yours. I was able to complete using an inner query, giving me a slightly
“cleaner” solution:
with bad_reviews AS (
select * from reviews
join places on
reviews.place_id = places.id
)
select username,count(*) from bad_reviews
where rating < (
select avg(average_rating)
from bad_reviews
)
group by username
order by count(*) desc
;
1
Reply
1 MONTH LATER
selectall
Good one i too did sth like the one u did.
username = @pinkdeb
review_count(bad) = 7
Reply
gr8 work just figuring out how could u think all this
Reply
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
Reply
8 DAYS LATER
selectall
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 Reply
1
Reply
selectallBeta Tester
marinsimeonoff346770
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.
1 Reply
5
Reply
selectall
That’s a valid point. “Average rating for places” is a little vague and is a question of interpretation. Thanks.
2
Reply
19 DAYS LATER
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.
1 Reply
Reply
valriecloutier078343
I put LIMIT 15 instead of 1 to validate my point, don’t get me wrong!
Reply
2 MONTHS LATER
hey guys here is what i did, let me know if makes sense
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;
Reply
29 DAYS LATER
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;
)
…
1 Reply
Reply
yizuhiProblem Solver (bronze)
gguirao13
Maybe it’s just that typo
1 Reply
1
Reply
yizuhi
Even after I corrected the spelling, I still don’t see anything.
1 Reply
Reply
yizuhiProblem Solver (bronze)
gguirao13
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
Reply
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;
2 Replies
Reply
19 DAYS LATER
raquelaurelia1999Today is the anniversary of the day I joined this community!
blog7064965569
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!
Reply
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
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;
Reply
WHERE reviews.rating < (SELECT AVG (average_rating) FROM places)
Why can’t I use:
WHERE reviews.rating < (AVG(places.average_rating))
instead?
I think there is no need to do average of average rating. I just did the initial inner join of places and reviews and compared the rating to the average rating.
WITH rev AS (SELECT name, average_rating, username, rating, review_date, note FROM places INNER JOIN reviews ON places.id=reviews.place_id) SELECT COUNT(*) AS “counts”, rev.username FROM rev WHERE rev.rating<rev.average_rating GROUP BY rev.username ORDER BY counts DESC;
This was my solution. I am not sure if it is correct or not though.
WITH bad_reviews AS (SELECT places.name, places.average_rating, reviews.username, reviews.rating, reviews.review_date FROM places JOIN reviews ON places.id = reviews.place_id WHERE reviews.rating < places.average_rating)
SELECT username, COUNT(rating) FROM bad_reviews GROUP BY username ORDER BY COUNT(rating) DESC LIMIT 1;
Would be thankful if somebody comment on my solution. I think we should choose simpler solution.
SELECT username, name, COUNT()
FROM reviews
CROSS JOIN places
ON places.id = reviews.place_id
WHERE rating < average_rating
GROUP BY username
ORDER BY COUNT() DESC
LIMIT 1;
This is what I did
Would love to hear your thoughts!
Hey gang!
I also got @pinkdeb. I couldn’t really see if really see anyone else used the same code I did. If someone has an eye for efficient code, would love if someone could let me know if I went about it in the best way. I did it this way so the result gave me a nice clean answer with two columns.
SELECT COUNT(reviews.username) AS ‘Number of Bad Reviews’, reviews.username
FROM reviews
LEFT JOIN places
ON reviews.place_id = places.id
WHERE reviews.rating < places.average_rating
GROUP BY reviews.username
ORDER BY 1 DESC
LIMIT 1;
Thanks so much!
Hi folks
Commenting here because I got a different answer than others have said
From what I found, @pinkdeb only has 7 reviews below average, but @its_ozzy had 8
Let me know if something is jumping out to say I have done it wrong and that’s why I have a different one
SELECT COUNT(r.username), r.username
FROM reviews AS r
JOIN places AS p
ON r.place_id = p.id
GROUP BY r.username
HAVING r.rating < p.average_rating
ORDER BY COUNT(r.username) DESC
LIMIT 1;