LEFT OUTER JOIN保留第一个表使用的所有匹配结果，而第二个表仅使用

Question

When I try to execute the mysql query below here I won't return all the record id's giving in the WHERE clause.

What I am trying to do:

1. Get all result records by id (IN)

2. JOIN the second table (websites_thumbs) By website_salt of Table 1 record result

   By salt of the record from table 1 but also a salt from the parent site ($parent_salt)

3. SUM votes and return in record result (if no result in the second table is found return 0)

ID 119 and 250 are having voting results in the second table, the others doesn't. The problem I got is when I execute the query it returns just the records 1,119 & 250 the other id's are existing but they are not returned?

The query:

// VxZQ85cByb98wUx0f3 => parent salt coming from another query (page data query)

SELECT 
    SUM(CASE t.thumb when 1 then 1 else 0 end) as thumbs_up,
    SUM(CASE t.thumb when -1 then 1 else 0 end) as thumbs_down,
    w.*, t.*
FROM websites as w
  LEFT OUTER JOIN websites_thumbs AS t 
  ON t.similar_website_salt = w.salt AND t.website_salt = 'VxZQ85cByb98wUx0f3' 
WHERE w.id IN ('1', '20', '31', '4', '199', '250', '633953')
GROUP BY t.similar_website_salt

first table

second table

Executing result

Hope someone can help me with this on!

Thanks

Answer 1

Instead of LEFT OUTER JOIN try INNER JOIN . Also, don't use ' in the IN clause. Lastly, I changed the order of your JOIN just because I'm picky :)

SELECT 
    SUM(CASE t.thumb when 1 then 1 else 0 end) as thumbs_up,
    SUM(CASE t.thumb when -1 then 1 else 0 end) as thumbs_down,
    w.*, t.*
FROM websites as w
  INNER JOIN websites_thumbs AS t 
  ON w.salt = t.similar_website_salt AND t.website_salt = 'VxZQ85cByb98wUx0f3' 
WHERE w.id IN (1, 20, 31, 4, 199, 250, 633953)
GROUP BY t.similar_website_salt

It's kind of hard to say without seeing table schema, so if you want to throw some of that up there it would be helpful.

Answer 2

Your use of GROUP BY is eliminating any rows with a null similar_website_salt . You need to change the null value to something else (eg empty string), and then group by that. Note there are two lines with the COALESCE addition:

SELECT 
    SUM(CASE t.thumb when 1 then 1 else 0 end) as thumbs_up,
    SUM(CASE t.thumb when -1 then 1 else 0 end) as thumbs_down,
    w.*, t.*,

    COALESCE(t.similar_website_salt, '')

FROM websites as w
  LEFT OUTER JOIN websites_thumbs AS t 
  ON t.similar_website_salt = w.salt AND t.website_salt = 'VxZQ85cByb98wUx0f3' 
WHERE w.id IN ('1', '20', '31', '4', '199', '250', '633953')

GROUP BY COALESCE(t.similar_website_salt, '')

Answer 3

Not all records are having a sub record in the second table so we can't GROUP BY t.similar_website_salt and keep records in the result set without this value.

SELECT 
    SUM(CASE t.thumb when 1 then 1 else 0 end) as thumbs_up,
    SUM(CASE t.thumb when -1 then 1 else 0 end) as thumbs_down,
    w.*, t.*
FROM websites as w
  LEFT OUTER JOIN websites_thumbs AS t 
  ON t.similar_website_salt = w.salt AND t.website_salt = 'VxZQ85cByb98wUx0f3' 
WHERE w.id IN ('1', '20', '31', '4', '199', '250', '633953')
//GROUP BY t.similar_website_salt
GROUP BY w.id <-- this does the trick