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mysql count 为null时，显示0
1.使用ifnull
ifnull（字段名，目标值）
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
select use_city,count(*) from data where os="Windows 2003 Std" group by use_city;
得到的结果为：
+-----------+----------+ | 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
+-----------+----------+ | 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：
SELECT use_city, COUNT(CASE WHEN os='Windows 2003 Std' THEN 1 ELSE NULL END) FROM data GROUP BY use_city
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