从一个MySQL表中选择，即使count为0，也返回count（id）

Question

I've been scouring google for the answer to this, and I'm SURE it's easier than I'm making it, and I'd like to learn the right way to do this.

I have one table emails_sent that I'm selecting counts from and grouping by a column. Each row has one of 3 type - "confirm", "1day", and "sameday". I'm also doing this by a certain date. So let's say, here's an example query I'm selecting for today:

SELECT Count(id) AS count, 
       type 
FROM   emails_sent 
WHERE  Date_format(date, '%Y-%m-%d') = '2013-03-29' 
GROUP  BY type 
ORDER  BY type; 

This is what it returns:

+-------+---------+
| count | type    |
+-------+---------+
|    32 | confirm |
|     4 | 1day    |
+-------+---------+

Which is great, but there is also a third "type" that happens to have 0 records for today, but I need it to pull that count even if it's 0.

I've also tried creating a 2nd table with just those 3 "types" in it, and using a right join, but it still won't pull the 0 count. This is what I did with the two tables:

SELECT t.type, 
       Count(s.id) AS count 
FROM   emails_sent s 
       RIGHT JOIN email_types t 
               ON s.type = t.type 
WHERE  Date_format(s.date, '%Y-%m-%d') = '2013-03-29' 
GROUP  BY t.type; 

I'm sure it's just something simple that I'm missing.

Answer 1

Try this

SELECT t.type, 
       COALESCE( Count(s.id), 0 ) AS count 
FROM   email_types t 
       LEFT JOIN email_sent s 
               ON s.type = t.type 
WHERE  Date_format(s.date, '%Y-%m-%d') = '2013-03-29' 
GROUP  BY t.type; 

Note: I have used a LEFT JOIN instead, as 1 type has many emails

Answer 2

如果我正确理解了该问题，您是否不能仅假设如果没有获得某种类型的记录，那是因为条目为0？