第三题包5%的，写前缀和像小丑

select f.farm_name as farm_name, c.crop_name, c.crop_type, round(avg(health_index), 2) as avg_health_index, round(sum(yield_kg_per_hectare),0) as avg_yield_kg_per_hectare, datediff(max(measurement_date), min(planting_date)) as growth_days from crop_data cd inner join farms f on cd.farm_id = f.farm_id inner join crops c on cd.crop_id = c.crop_id where cd.measurement_date between '2024-03-01' and '2024-06-30' group by cd.crop_id, cd.farm_id,f.farm_name having count(cd.data_id) >= 3 order by avg_health_index desc, avg_yield_kg_per_hectare desc,farm_name

我想问下面试java能不能用c/c++写编程题

Java让我转交付工程师了 我是废物啊 第三题A不出来

第一题SQL有个坑，求平均公顷，我看他案例，明明求的是SUM，换成SUM就过了

兄弟，一样

第二道咋A的

同学，瞅瞅我司，校招刚开，点击就送，我的主页最新动态，绿灯直达

无语

我感觉第3题可以直接使用Java的BigInteger去逃课，但是我仍然没写出来，依然是5%

第二道咋A的

select f.farm_name, c.crop_name, c.crop_type, round(sum(d.health_index) / count(c.crop_name), 2) avg_health_index, round(max(d.yield_kg_per_hectare)) avg_yield_kg_per_hectare, datediff(max(d.measurement_date), min(d.planting_date)) growth_days from crop_data d join crops c on d.crop_id = c.crop_id join farms f on d.farm_id = f.farm_id where d.measurement_date between '2024-03-01' and '2024-06-30' group by f.farm_name, c.crop_name, c.crop_type, c.crop_name having count(d.measurement_date) >= 3 order by avg_health_index desc, avg_yield_kg_per_hectare desc, f.farm_name;

我也是一三没搞出来

数据库数据格式转换函数不会，第三题改出来一个bug之后从5%到0%了，麻了

虽然我很菜，但是不管你运维还是开发，数据库都是基础，尤其是运维，而且面试必考

sql题在我idea上运行的好好的，放测试环境报错了

数据库不知道哪里有问题，平均公顷数那一列一直错。第三题暴力A了5%

第三题什么思路，乘几下就越界了long