在我正在处理的数据库中,有14个不同城市的新闻表(table1,table2 .... table14)具有完全相同的结构。一些专栏是: id,作者,日期,news_title,news_body
有不同的作者在相关城市表中发布。现在我想知道今天每个作者在所有14个表格中发布了多少新闻。
以下是输出格式:
author | num_of_post
bob 5
peter 12
请记住,有14个表(这是问题)
答案 0 :(得分:1)
正如@Grodon已经说过你可以使用这样的联盟:
select author, count(id) as num_of_post from table1 group by author
union
select author, count(id) as num_of_post from table2 group by author
union
select author, count(id) as num_of_post from table3 group by author
等等
更新
由于您有多个表中的1位作者的记录,因此您可以再次对联合结果进行分组。
select author, sum(num_of_post) from (
select author, count(id) as num_of_post from table1 group by author
union
select author, count(id) as num_of_post from table2 group by author
union
select author, count(id) as num_of_post from table3 group by author) a group by author