2023-05-02 01:27:09
1、
select car.numplate as license ,COUNT( pet.petid) pet_count from pet
left join person on person .perid =pet.perid
left join car on person .perid =car.perid
group by car. numplate
2、
select person.sex gender ,COUNT( person.perid) 1t100000_count from person
where income<100000
3、

第三个题的答案粘贴文本删帖不了,附上图片。
2023-09-13 21:01:27
2022-07-27 09:35:08
好像每次写代码百度都会抽掉,我还是用图片放出我的答案吧,以下原答案:

2022-06-14 13:42:55

好心人在此···