Question

我正在为iOS应用程序提供所有php服务。其中一个似乎非常快地消耗我所有的3个CPU内核（而没有其他内容）。该服务将51个最近的“气泡”（bulles）提取到用户实际所在的位置。这是我的php：

$sql="SELECT *, get_distance('$latitude','$longitude',latitude,longitude) AS proximite
FROM bulles INNER JOIN
(SELECT userId, userFirstName, userLastName, userName, userPictureUrl from user) AS users
ON bulles.auteur = users.userId
INNER JOIN interests
ON bulles.interet1 = interests.id
WHERE auteur = '$userID'
OR auteur
IN (
    SELECT followedId
    FROM following
    WHERE followerId = '$userID'
)
ORDER BY proximite
LIMIT 51";

这是我存储的函数“get_distance”：

FUNCTION `get_distance`(lat1 DOUBLE, lng1 DOUBLE, lat2 DOUBLE, lng2 DOUBLE) RETURNS double
BEGIN
    DECLARE rlo1 DOUBLE;
    DECLARE rla1 DOUBLE;
    DECLARE rlo2 DOUBLE;
    DECLARE rla2 DOUBLE;
    DECLARE dlo DOUBLE;
    DECLARE dla DOUBLE;
    DECLARE a DOUBLE;

    SET rlo1 = RADIANS(lng1);
    SET rla1 = RADIANS(lat1);
    SET rlo2 = RADIANS(lng2);
    SET rla2 = RADIANS(lat2);
    SET dlo = (rlo2 - rlo1) / 2;
    SET dla = (rla2 - rla1) / 2;
    SET a = SIN(dla) * SIN(dla) + COS(rla1) * COS(rla2) * SIN(dlo) * SIN(dlo);
    RETURN (6378137 * 2 * ATAN2(SQRT(a), SQRT(1 - a)));
END

每秒10次此服务请求使cpu饱和。我之前遇到过INNER JOIN语句的问题，所以也许它来自那里？有什么想法吗？

Answer 1

MySQL'IN'效率非常低（虽然是一个很好的构造），因为查询会针对每一行进行重新处理。即。

IN (
    SELECT followedId
    FROM following
    WHERE followerId = '$userID'
)

将其替换为FROM子句中的表。 e.g：

...FROM ( SELECT followedId FROM following WHERE followerId = '$userID') AS whatever 
(then in the WHERE put ) OR auteur = whatever.followedId...

干杯。

使用php的高CPU消耗Mysql请求

1 个答案: