mysql count 为null时,显示0
1.使用ifnull
ifnull(字段名,目标值)
1
2
3
4
5
|
SELECT a.*,IFNULL(r. count ,0) from ( SELECT act_id,poster_id,type,style_type,status,status_time,title,content,images,start_time,end_time,district_id,address,lon_map,lat_map,person_num,person_name,person_phone,person_wx_id,reward_min,reward_max,cost_type,meal_svc,taxi_svc,hair_length,hair_handle,remark,is_over,create_time,note FROM activity WHERE poster_id = 3055808629673784641 ORDER BY create_time DESC limit 0,20) a LEFT JOIN ( SELECT count (1) count ,act_id FROM activity_user u where u.user_status in ( 'YBM' , 'YTG' , 'YQR' ) GROUP BY u.act_id) r on a.act_id = r.act_id |
2.运行结果
mysql让count为0的记录也显示出来
在mysql 下执行如下命令
1
|
select use_city, count (*) from data where os= "Windows 2003 Std" group by use_city; |
得到的结果为:
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
|
+ -----------+----------+ | use_city | count (*) | + -----------+----------+ | Beijing | 2 | | Chengdu | 2 | | Chongqing | 1 | | Dalian | 2 | | Fuzhou | 2 | | Guangzhou | 2 | | Hangzhou | 2 | | Nanjing | 2 | | Shanghai | 21 | | Shenyang | 5 | | Wuhan | 1 | + -----------+----------+ 11 rows in set (0.01 sec) |
共有11条记录,问题是其中有一个城市“Tianjing”是没有记录符合的,怎么让他显示成
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
|
+ -----------+----------+ | use_city | count (*) | + -----------+----------+ | Beijing | 2 | | Chengdu | 2 | | Chongqing | 1 | | Dalian | 2 | | Fuzhou | 2 | | Guangzhou | 2 | | Hangzhou | 2 | | Nanjing | 2 | | Shanghai | 21 | | Tianjing | 0 | | Shenyang | 5 | | Wuhan | 1 | + -----------+----------+ |
让它count为0的记录也显示出来,做法如下:
1
2
3
4
5
6
7
|
SELECT use_city, COUNT ( CASE WHEN os= 'Windows 2003 Std' THEN 1 ELSE NULL END ) FROM data GROUP BY use_city |
以上为个人经验,希望能给大家一个参考,也希望大家多多支持服务器之家。
原文链接:https://blog.csdn.net/weixin_37460672/article/details/119273481