恒为艺术-记录程序学习的点点滴滴

SELECT COUNT(*) FROM test_users WHERE 1 = 1 GROUP BY email,passwords COUNT(*) 1 2 1 1 显然这不是我要的结果，这样统计出来的是相同email和passwords的各个记录数量之和,下面这样就可以了 SELECT COUNT(DISTINCT email,passwords) F...

show full processlist; kill 进程号;