Question

我正在开发一个Web项目，其中涉及从mysql获取一些计算值。我已经编写了一个执行此操作的sql查询，但它是一个子查询，它使应用程序运行速度大大降低。现在，我尝试使用在线阅读的 join 优化查询。我是编程概念的新手，尤其是mysql，我希望有人指出正确的路径。

当我尝试使用SUM和CASE获取mysql中值的总和时，我得到了一个不正确的值（超过值）。

以下是查询：

SELECT 
master.lgName,
master.wardName,
master.pUnitName,
SUM(voter_reg_no) AS vregno,
pvc_collected,
COUNT(DISTINCT master.wardName) as no_ward,
COUNT(DISTINCT CASE WHEN master.reportedpu = '1' THEN master.reportedpu END) as reportedpu,
SUM(CASE WHEN total != '' AND master.lgName = accreditation.lga_name THEN total ELSE 0 END) as acr,
SUM(CASE WHEN category = 'ivc'  THEN cno ELSE 0 END) AS ivc,
SUM(CASE WHEN category = 'tvc'  THEN cno ELSE 0 END) AS tvc,
SUM(CASE WHEN category = 'other'  THEN cno ELSE 0 END) AS oth,
SUM(CASE WHEN category = 'apc'  THEN cno ELSE 0 END) AS apc,
SUM(CASE WHEN category = 'pdp' THEN cno ELSE 0 END) AS pdp,
SUM(CASE WHEN category = 'adp' THEN cno ELSE 0 END) AS adp,
SUM(CASE WHEN category = 'adc'  THEN cno ELSE 0 END) AS adc,
SUM(CASE WHEN category = 'ad'  THEN cno ELSE 0 END) AS ad,
SUM(CASE WHEN category = 'sdp'  THEN cno ELSE 0 END) AS sdp
FROM master LEFT JOIN res ON master.lgName = res.lgName 
LEFT JOIN accreditation ON accreditation.lga_name = master.lgName
group by master.lgName ORDER by master.userCode ASC

下面是我之前的查询，该查询为我提供了正确的值，但降低了应用程序的速度：

            SELECT a.lgName as lgname, a.wardName as wardName, a.pUnitName as pUnitName,sum(a.pvc_collected) as pvc, sum(a.voter_reg_no) as purvs, COUNT(DISTINCT wardName) as No_wards, 
count(pUnitName) as pu, 
            (select count(reportedpu) from master where wardName = a.wardName and reportedpu = 1) as reportedpu,
            (SELECT sum(total) FROM accreditation WHERE ward_name IN (select wardName from master where wardName = a.wardName)) AS acr, 
            (SELECT sum(cno) FROM res WHERE wardName IN (select wardName from master where wardName = a.wardName) and category = 'tvc') AS tvc,
    (SELECT sum(cno) FROM res WHERE wardName IN (select wardName from master where wardName = a.wardName) and category = 'apc') AS apc,
    (SELECT sum(cno) FROM res WHERE wardName IN (select wardName from master where wardName = a.wardName) and category = 'ivc') AS ivc,
    (SELECT sum(cno) FROM res WHERE wardName IN (select wardName from master where wardName = a.wardName) and category = 'pdp') AS pdp,
    (SELECT sum(cno) FROM res WHERE wardName IN (select wardName from master where wardName = a.wardName) and category = 'ad') AS ad,
    (SELECT sum(cno) FROM res WHERE wardName IN (select wardName from master where wardName = a.wardName) and category = 'adc') AS adc,
     (SELECT sum(cno) FROM res WHERE wardName IN (select wardName from master where wardName = a.wardName) and category = 'sdp') AS sdp,
      (SELECT sum(cno) FROM res WHERE wardName IN (select wardName from master where wardName = a.wardName) and category = 'adp') AS adp,

    (SELECT sum(cno) FROM res WHERE wardName IN (select wardName from master where wardName = a.wardName) and category = 'other') AS oth
    FROM master a group by a.lgName order by a.userCode asc

使用CASE对mysql中的行求和时获取值

0 个答案: