如果在mysql查询中找不到任何结果，填充空白数据

Question

我有一个查询正在检查5个评论网站的数据并返回site_id，review_count & review_average。

如果没有评论网站的数据，那么我想返回0为&的平均值。

这是可能在一个MySQL查询吗？

MySQL的：

SELECT rrss.review_site_id,rrss.review_count,rrss.review_average,rs.name 
FROM rooftops_review_sites_snapshots rrss 
LEFT JOIN review_sites rs ON rrss.review_site_id = rs.id 
WHERE rrss.rooftop_id = 185 
AND rrss.import_id = 16 
AND rrss.review_site_id IN (31,30,12,10,29) 

电流输出：

Array 
(
[google] => Array 
    (
     [review_site_id] => 31 
     [review_count] => 24 
     [review_average] => 3.80 
    ) 

[edmunds] => Array 
    (
     [review_site_id] => 12 
     [review_count] => 8 
     [review_average] => 4.50 
    ) 

) 

所需的输出：

Array 
(
[google] => Array 
    (
     [review_site_id] => 31 
     [review_count] => 24 
     [review_average] => 3.80 
    ) 

[edmunds] => Array 
    (
     [review_site_id] => 12 
     [review_count] => 8 
     [review_average] => 4.50 
    ) 
[yelp] => Array 
    (
     [review_site_id] => 31 
     [review_count] => 0 
     [review_average] => 0 
    ) 

[dealerrater] => Array 
    (
     [review_site_id] => 12 
     [review_count] => 0 
     [review_average] => 0 
    ) 
[cars] => Array 
    (
     [review_site_id] => 12 
     [review_count] => 0 
     [review_average] => 0 
    ) 

) 

Answer 1

试试这个

SELECT 
rrss.review_site_id, 
ISNULL(rrss.review_count, 0), 
ISNULL(rrss.review_average, 0), 
rs.name 

FROM rooftops_review_sites_snapshots rrss 

LEFT OUTER JOIN review_sites rs 
ON rrss.review_site_id = rs.id 

WHERE rrss.rooftop_id = 185 
AND rrss.import_id = 16 
AND rrss.review_site_id IN (31,30,12,10,29) 

Answer 2

你的表的更换位置和使用ISNULL（）。检查下面的查询。

SELECT rs.id,rs.name, ISNULL(rrss.review_count, 0), ISNULL(rrss.review_average, 0) FROM review_sites rs LEFT JOIN rooftops_review_sites_snapshots rrss ON rrss.review_site_id = rs.id WHERE rrss.rooftop_id = 185 AND rrss.import_id = 16 AND rrss.review_site_id IN (31,30,12,10,29)