$query="SELECT count(queries.id) as queries, count(answers.id) as answers, count(users.id) as users FROM queries,answers,users";不对,我试了!!
不可以! 你 FROM queries,answers,users"; 同时打开3张表但没有给出他们之间的连接关系,数据库系统将假定他们全连接,只能得到3张表记录数的乘积只能 select count(*) as cnt FROM queries union select count(*) as cnt FROM answers union select count(*) as cnt FROM users;将得到3条记录的返回结果,分别为3张表的记录数
$query="SELECT count(*) AS queries_sum FROM queries union SELECT count(*) AS answers_sum FROM answers union SELECT count(*) AS users_sum FROM users"; 还有错!
你 FROM queries,answers,users";
同时打开3张表但没有给出他们之间的连接关系,数据库系统将假定他们全连接,只能得到3张表记录数的乘积只能
select count(*) as cnt FROM queries
union
select count(*) as cnt FROM answers
union
select count(*) as cnt FROM users;将得到3条记录的返回结果,分别为3张表的记录数
是的,楼上您的理解正确!
我的到的结果是:queries.id*answers.id*users.id=280(即5*8*7)
还有错!