计数2列
这是我的桌子
Table A
|----------------------|---------------------|------------------|
| id | typec NOM |
|----------------------|---------------------|------------------|
| 1 | Plat | A |
|----------------------|---------------------|------------------|
| 2 | Steeple-chase | B |
|----------------------|---------------------|------------------|
| 3 | Haies | C |
|----------------------|---------------------|------------------|
我希望输出为:
+------------+--------+--------------+---------+-----------+
| nom | Plat |Steeple-chase | Haies | ALL |
+------------+--------+--------------+---------+-----------+
| A | 10 | 2 | 8 | 20 |
+------------+--------+--------------+---------+-----------+
| B | 100 | 50 | 50 | 100 |
+------------+--------+--------------+------+--------------+
| C | 5 | 5 | 5 | 15 |
+------------+--------+--------------+---------+-----------+
目前，我运行此查询，但结果不是很好
SELECT nom, typec, count(*) AS count
FROM p_mu.cachedate
GROUP BY nom, typec;
谢谢你的帮助。。。
我怀疑您正在寻找枢轴/条件聚合（如果您发布了与期望结果匹配的示例数据，我会更加确定）
SELECT nom, sum(case when typec ='plat' then 1 else 0 end) as 'plat',
sum(case when typec ='steeple_chase' then 1 else 0 end) as 'steeple-chase',
sum(case when typec ='hais' then 1 else 0 end) as 'hais',
count(*) AS 'all'
FROM p_mu.cachedate
GROUP BY nom;
您要查找的函数不是COUNT（），它是+运算符
SELECT nom, typec1, typec2, typec3, typec1+typec2+typec3 AS ALL
FROM p_mu.cachedate
GROUP BY nom
试试看，如果这是有效的，并分享结果！
