Question

我有一个看起来像这样的SQL表，但是有数十万行：

+------+---------------------+-----------+------------+--------------+----------+
| id   | timestamp           | lat       | lon        | country_code | city     |
+------+---------------------+-----------+------------+--------------+----------+
| 2231 | 2013-09-22 14:58:32 | 28.179199 | 113.113602 | CN           | Changsha |
| 2232 | 2013-09-22 14:58:32 | 28.179199 | 113.113602 | CN           | Changsha |
| 2233 | 2013-09-22 14:58:32 | 41.792198 | 123.432800 | CN           | Shenyang |
| 2234 | 2013-09-22 14:58:32 | 31.045601 | 121.399696 | CN           | Shanghai |
| 2235 | 2013-09-22 14:58:32 | 45.750000 | 126.650002 | CN           | Harbin   |
| 2236 | 2013-09-22 14:58:32 | 39.928902 | 116.388298 | CN           | Beijing  |
| 2237 | 2013-09-22 14:58:32 | 26.061399 | 119.306099 | CN           | Fuzhou   |
| 2238 | 2013-09-22 14:58:32 | 26.583300 | 106.716698 | CN           | Guiyang  |
| 2239 | 2013-09-22 14:58:32 | 39.928902 | 116.388298 | CN           | Beijing  |
| 2240 | 2013-09-22 14:58:32 | 31.045601 | 121.399696 | CN           | Shanghai |
+------+---------------------+-----------+------------+--------------+----------+

我需要根据时间戳（间隔）进行查询并获取适合该时间间隔的所有记录，计算具有相同城市的行（并添加任何项目的纬度/经度，它们将完全相同同一组）。目前我的应用程序代码中只有一个普通的select和组（如下所示），但这很慢，因为它需要向应用程序发送几百kb。

（聚合的python代码）

sorted_events = sorted(result, key=itemgetter('city'), reverse=False)
    for k, g in groupby(sorted_events, key=itemgetter('city')):
        group = list(g)
        first_item = group[0]
        unique_city_item = { 
            "city" : first_item['city'], 
            "country_code" : first_item['cc'], 
            "lon" : first_item['lon'], 
            "lat" : first_item['lat'], 
            "number_of_items" : len(group)
        }

它按照我想要的方式工作，但速度很慢。有没有办法直接用SQL查询执行此操作？我得到以下JSON输出，我想要类似的东西：

{
    {
        city: "Baotou",
        lon: 109.822197,
        country_code: "CN",
        lat: 40.652199,
        number_of_items: 288
    },
    {
        city: "Beijing,",
        lon: 116.388298,
        country_code: "CN",
        lat: 39.928902,
        number_of_items: 47
    }
}

Answer 1

这是你在找什么？

select city, lon, country_code, lat, count(*) as number_of_items
from table t
where timestamp between STARTTIMESTAMP and ENDTIMESTAMP
group by city, lon, country_code, lat;

group mysql按字段查询具有相同的值和计数

1 个答案: