我有下表:
topic_id(unique) forum_id forum_views
1002 1885 5
1003 1893 2
1004 1885 3
1005 1892 6
我如何得到这样的输出:(独特的forum_id,上面的论坛总观点次数)
forum_id forum_views
1885 8
1893 2
1892 6
任何帮助都得到了赞赏。
这就是我想要的:
从table_name
中选择不同的forum_id,sum(topic_views)
答案 0 :(得分:7)
select forum_id,
sum(forum_views) as forum_views
from jforum_topics
group by forum_id
答案 1 :(得分:2)
使用此查询select forum_id, sum(distinct forum_views) from table group by forum_id;
在此查询中, sum distinct 将执行不同值的总和,并按所需列进行分组。
如果您想要所有值的总和而不考虑清晰度,那么只需使用sum(forum_views)
。
答案 2 :(得分:1)
select forum_id, sum(forum_views) as sum from table_name group by forum_id
答案 3 :(得分:0)
您只需要GROUP BY
来指定您希望SUM
应用于哪些行;
select forum_id, sum(topic_views) from jforum_topics
group by forum_id
order by forum_id
答案 4 :(得分:0)
试试这个:
select forum_id, sum(topic_views)
from jforum_topics
group by forum_id
答案 5 :(得分:0)
select forum_id, sum(topic_views) from jforum_topics group by forum_id
答案 6 :(得分:0)
试试这个,它会将你的forum_views与forum_id相加。
SELECT forum_id
, SUM(forum_views)
FROM TABLE_NAME
GROUP BY forum_id
ORDER BY 1
答案 7 :(得分:0)
它将是
select id2,sum(id3) from supportContacts group by id2
请参阅此SQL小提琴http://sqlfiddle.com/#!2/7b9e7/3
答案 8 :(得分:0)
你也可以试试:
select distinct forum_id, sum(topic_views)
from jforum_topics
group by forum_id