Question

我的查询正在检查5个评论网站的数据并返回site_id，review_count＆amp; review_average。

如果评论网站没有数据，那么我想返回0作为计数＆amp;平均。

这可以在mysql查询中执行吗？

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)

当前输出：

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
    )

)

期望的输出：

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

更改表的位置并使用ISNULL（）。查看下面的查询。

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)

Answer 2

尝试这个

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)

如果在mysql查询中找不到结果，请填写空白数据

2 个答案: