我有一个大的SQL表,它存储了用户添加到他们的配置文件中的添加项目id和他们给它的评分。问题是,我需要快速计算一个项目添加了多少次,还需要获得它的总评分。有时,评级可能为空,也可能不为空,这取决于用户是否为项目留下了评级。这张桌子看起来有点像
Item_Feedback_ID (Pri, AI) Item ID Item Rating User_ID_Added
1 5 Null 4
2 5 8 5
3 6 9 9
我需要为每个唯一的项目ID做这件事(我已经有了一个存储唯一项目ID的表)。我现在正在做的是:
PSEUDO PHP CODE:
$result = the result of: "SELECT item_id FROM items";
foreach ($item_id in $result) {
$sql = "SELECT COUNT(item_id) as sum, SUM(`Item Rating`) as total_rating FROM
item_feedback WHERE item_id = $item_id";
//Run that sql statement in PHP and parse it and save in an array
}
有更有效的方法吗?我似乎浪费了很多时间,因为COUNT已经扫描了所有的列,我必须再次扫描x个项目ID。
您应该能够使用可选数据的子查询来完成这一查询。我对额外的数据使用了LEFT JOIN,这样没有这些数据的项仍然会被返回。子查询的原因是您使用的是COUNT
。如果没有子查询,这将计算所有记录,而不仅仅是有反馈的记录。
SELECT
items.item_id,
feedback.sum,
feedback.total_rating
FROM items
LEFT JOIN(
SELECT
item_id,
COUNT(item_id) as sum,
SUM(`Item Rating`) as total_rating
FROM item_feedback
GROUP BY item_id
) AS feedback ON feedback.item_id = items.item_id
我不是PHP专家,但也许你可以把数据放入一个临时表中,然后循环使用它。
INSERT INTO TempTable
SELECT
COUNT(item_id) as sum,
SUM(`Item Rating`) as total_rating
FROM item_feedback
GROUP BY item_id
然后通过item_id 循环TempTable
$sql = "SELECT ifnull(t.count, 0) as count,
ifnull(t.total_rating, 0) total_rating,
a.item_id
FROM items a
left join (SELECT COUNT(item_id) as count,
SUM(`Item Rating`) as total_rating,
b.item_id
from item_feedback b
group by b.item_id) t on t.item_id = a.item_id " ;
如果您不需要每个Item都有一行,那么只需要查询Item_feedback就足够了,使用SQL CASE
语句也会帮助您:
select
item_id,
count(*) as item_added_count,
sum(case when [item rating] is not null then 1 else 0 end) as rating_count
from item_feedback
group by item_id
如果每个项目都需要一行,并且希望在项目从未添加到item_feedback表中时显示0,则对项目表使用LEFT OUTER JOIN
:
select
i.item_id,
sum(case when f.item_id is not null then 1 else 0 end) as item_added_count,
sum(case when f.[item rating] is not null then 1 else 0 end) as rating_count
from items i
left outer join item_feedback f on f.item_id = i.item_id
group by i.item_id