問(wèn)題描述

問(wèn)題解答

select j.*, (select count(*) from submitjobs s where j.id=s.jobid) from jobs j回答2：

mysql 為例：select a.*,count(b.userid)sbtotal from jobs a left join sumbmitjobs b on b.jobid=a.id gourp by b.jobid order b.userid

oracle 差不多 只是不能那樣分組需要先分組sumbmitjobs 統(tǒng)計(jì)好然后作為子查詢(xún)回答3：

SELECT COUNT(DISTINCT column(jobName)) FROM jobs LEFT JOIN submitjobs ON jobs.jobName=submitjobs.jobName

select b.name, a.cnt from jobs b left join (select jobid, count(userid) as cntfrom submitjobsgroup by jobid) aon b.id = a.jobid回答5：

沒(méi)測(cè)試...

SELECT j.id, j.name, COUNT(sj.jobid) AS countFROM `submitjobs`AS sjJOIN `jobs`AS jWHERE sj.jobid = j.idGROUP BY sj.jobid回答6：

其實(shí)最簡(jiǎn)單

select 想要的字段 from 表1,表2 ...表N

此種寫(xiě)法不管有多少表都能連起來(lái)