我有一个如下的 MySQL 函数,但它正在生成错误:
BEGIN
DECLARE image1 VARCHAR(250);
select case when
(
select COUNT(*)
from profile_images
where building_id = bid
and contractor_id = cid
) > 0
then (
select distinct (image)
-- into image1
from profile_images
where building_id = bid
and contractor_id = cid limit 1
) else (
select distinct (image)
-- into image1
from profile_images
where contractor_id = cid limit 1
)
END into image1;
RETURN image1;
END
MySQL 显示的实际错误是 #2014 - Commands out of sync; 你现在不能运行这个命令。