php - 如果mysql查询没有结果则填写空白数据

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 和 review_average。

如果评论网站没有数据，那么我想返回 0 作为计数和平均值。

这可以在 mysql 查询中完成吗？

mySQL:

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

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)