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我有一個查詢正在檢查5個評論網站的數據並返回site_id,review_count & review_average。如果在mysql查詢中找不到任何結果,填充空白數據
如果沒有評論網站的數據,那麼我想返回0爲&的平均值。
這是可能在一個MySQL查詢嗎?
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)
電流輸出:
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
)
)
所需的輸出:
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
)
)
如果'review_sites'表中存在的所有站點,它切換到謂語用另一臺在'LEFT JOIN'。 – 2014-10-30 00:01:52