mysql下的多一表多结果的联合查询外加求和

今天在开发过程中，碰到一难题，通过代码可以简单解决，但是考虑到量多时的效率，想通过sql解决下：

题目如何：

有两张表，通过用户id关联，

table1:

CREATE TABLE `tbusiness_evaluation_report` (

  `id` int(11) NOT NULL AUTO_INCREMENT,

  `userid` int(11) DEFAULT NULL,

  `loanrequestid` int(11) DEFAULT NULL,

  `code` varchar(45) CHARACTER SET utf8 DEFAULT NULL,

  `amount` decimal(10,2) DEFAULT '0.00',

  `goodstype` varchar(50) COLLATE utf8_bin DEFAULT NULL,

  `createtime` datetime DEFAULT CURRENT_TIMESTAMP,

  `desc` varchar(500) CHARACTER SET utf8 DEFAULT NULL,

  `rate` decimal(2,2) DEFAULT '0.00',

  `rateamount` decimal(10,2) DEFAULT '0.00',

  `interestrate` float DEFAULT '0',

  `interestamount` decimal(10,2) DEFAULT NULL,

  `state` int(11) DEFAULT '1',

  `title` varchar(45) CHARACTER SET utf8 DEFAULT NULL,

  `lockamount` decimal(10,2) DEFAULT '0.00',

  `moneystate` int(11) DEFAULT '0',

  `state1time` datetime DEFAULT NULL,

  `state2time` datetime DEFAULT NULL,

  `state3time` datetime DEFAULT NULL,

  `state1rate` float DEFAULT '0',

  `state2rate` float DEFAULT '0',

  `state3rate` float DEFAULT '0',

  `state1amount` decimal(10,2) DEFAULT '0.00',

  `state2amount` decimal(10,2) DEFAULT '0.00',

  `state3amount` decimal(10,2) DEFAULT '0.00',

  `matchendtime` datetime DEFAULT NULL,

  `state1fund` decimal(10,2) DEFAULT '0.00',

  `state2fund` decimal(10,2) DEFAULT '0.00',

  `state3fund` decimal(10,2) DEFAULT '0.00',

  `inviteid` int(11) DEFAULT '0',

  `platformuseamount1` decimal(10,2) DEFAULT '0.00',

  `platformuseamount2` decimal(10,2) DEFAULT '0.00',

  `state3userid` int(11) DEFAULT '0',

  PRIMARY KEY (`id`)

) ENGINE=InnoDB AUTO_INCREMENT=180 DEFAULT CHARSET=utf8 COLLATE=utf8_bin;

table2:

DROP TABLE IF EXISTS `tbusiness_pawn_ticket`;

CREATE TABLE `tbusiness_pawn_ticket` (

  `id` int(11) NOT NULL AUTO_INCREMENT,

  `code` varchar(50) DEFAULT NULL,

  `state` int(11) DEFAULT '1',

  `starttime` datetime DEFAULT CURRENT_TIMESTAMP,

  `endtime` datetime DEFAULT NULL,

  `pawnid` int(11) DEFAULT '0',

  `reportid` int(11) DEFAULT '0',

  `requestid` int(11) DEFAULT '0',

  `relevantid` int(11) DEFAULT '0',

  `relevanttype` int(1) DEFAULT '0',

  `overdueinterest` decimal(10,2) DEFAULT '0.00',

  `title` varchar(45) DEFAULT NULL,

  `term` int(11) DEFAULT '30',

  `userid` int(11) NOT NULL DEFAULT '0',

  `ransomtime` datetime DEFAULT NULL,

  `deadpawntime` datetime DEFAULT NULL,

  `createtime` datetime DEFAULT CURRENT_TIMESTAMP,

  `continuestate` int(11) DEFAULT '0',

  `continuecount` int(11) DEFAULT '0',

  PRIMARY KEY (`id`)

) ENGINE=InnoDB AUTO_INCREMENT=13 DEFAULT CHARSET=utf8;

在table1中有三个不同字段来表示用户ID（userid，inviteid，state3userid）,且这三个字段的值不会相等，

那就是说通过一个ID获取到userid，nviteid，state3userid等于这个ID的结果值有三个集合，再求和这三个集合不同字段的值的和

方法有三种：

最简单的是方法1：

select b.interestamount,b.state1amount,b.state1rate,
b.state2amount,b.state2rate,b.state3amount,b.state3rate,
b.platformuseamount2,b.userid,b.inviteid,b.state3userid
FROM tbusiness_pawn_ticket a LEFT JOIN tbusiness_evaluation_report b on a.reportid = b.id 
where (b.userid = userid or b.inviteid= userid or b.state3userid = userid) 
and a.state in (1,3)

然后再通过程序来计算值。

但是感觉量一上去拖慢速度，几百几千条时呢

打算想想是否可以一句sql解决，一开始写的时候一直找不到感觉，写不出来，打算曲线救国，写个储存过程吧：

所以有了方法2：

DROP PROCEDURE IF EXISTS calc_dstx; -- 如果存在此名的存储过程，先删除

CREATE PROCEDURE calc_dstx(IN userid INT,OUT amount FLOAT) -- 创建名为store_procedure的存储过程 待收本息

BEGIN -- 开始
DECLARE done INT DEFAULT -1;
DECLARE temp FLOAT DEFAULT 0;
DECLARE t_interestamount FLOAT DEFAULT 0;

  DECLARE t_state1amount FLOAT DEFAULT 0;
DECLARE
t_state1rate FLOAT DEFAULT 0;
DECLARE t_state2amount FLOAT DEFAULT 0;
DECLARE t_state2rate FLOAT DEFAULT 0;  
DECLARE
t_state3amount FLOAT DEFAULT 0;
DECLARE t_state3rate FLOAT DEFAULT 0;
DECLARE t_platformuseamount2 FLOAT DEFAULT 0;  
DECLARE t_userid INT DEFAULT 0;  
DECLARE t_inviteid INT DEFAULT 0;  
DECLARE t_state3userid INT DEFAULT 0;  
DECLARE report CURSOR FOR select b.interestamount,b.state1amount,b.state1rate,
b.state2amount,b.state2rate,b.state3amount,b.state3rate,
b.platformuseamount2,b.userid,b.inviteid,b.state3userid
FROM tbusiness_pawn_ticket a LEFT JOIN tbusiness_evaluation_report b on a.reportid = b.id 
where (b.userid = userid or b.inviteid= userid or b.state3userid = userid) 
and a.state in (1,3);
DECLARE CONTINUE HANDLER FOR NOT FOUND SET done=1; -- 如果没有记录，错误处理no data to fetch条件，几乎所有的游标都要用到
SET done=0;
SET temp=0;
OPEN report;
my_loop: LOOP
FETCH report INTO t_interestamount,t_state1amount,t_state1rate,t_state2amount,t_state2rate,t_state3amount,t_state3rate,t_platformuseamount2,t_userid,t_inviteid,t_state3userid;
IF done=1 THEN
LEAVE my_loop;
END IF;
IF userid=t_userid THEN

SET temp = temp + t_state1amount + t_interestamount*t_state1rate - t_platformuseamount2*t_state1rate;
END IF;
IF userid = t_inviteid THEN
SET temp = temp + t_state2amount+t_interestamount*t_state2rate - t_platformuseamount2*t_state2rate;
END IF;
IF userid = t_state3userid THEN
SET temp = temp + t_state3amount+t_interestamount*t_state3rate - t_platformuseamount2*t_state3rate;
END IF;
END LOOP my_loop;
CLOSE report;
-- DEALLOCATE report;
SET amount=temp;

END; -- 结束

-- DEMILITER ; -- 恢复;为分隔符

写完之后自信心爆棚，灵感爆发，一口气写出了第三种方法，一条sql

方法3：

select FORMAT(sum(money),2) as total from (

select b.userid as moneyuserid,(b.state1amount + b.interestamount*b.state1rate - b.platformuseamount2*b.state1rate) as money from tbusiness_pawn_ticket a LEFT JOIN tbusiness_evaluation_report b on a.reportid = b.id 
where b.userid = 1080
and a.state in (1,3)

UNION ALL select d.inviteid as moneyuserid,(d.state2amount + d.interestamount*d.state2rate - d.platformuseamount2*d.state2rate) as money from tbusiness_pawn_ticket c LEFT JOIN tbusiness_evaluation_report d on c.reportid = d.id 
where d.inviteid = 1080
and c.state in (1,3)

UNION ALL select f.state3userid as moneyuserid,(f.state3amount + f.interestamount*f.state3rate - f.platformuseamount2*f.state3rate) as money from tbusiness_pawn_ticket e LEFT JOIN tbusiness_evaluation_report f on e.reportid = f.id 
where f.state3userid = 1080 and e.state in (1,3)) as totaltabale

途中碰到几个问题，关于储存过程的循环跳出，关于mysql的全连接需要使用union all，关于集合做表查询必须有个别名。

种种还是基础不够扎实。

先记录下这个，对于这三种效率问题，还没有深入研究，等空余时间深入研究下