Mysql 从多个表中选择时选择sum

Mysql 从多个表中选择时选择sum,mysql,Mysql,我正在尝试选择一些值,我使用了以下查询: SELECT l.*,SUM(lp.amount) AS landPayMonth,p.pName,u.uName,SUM(ci.amount) AS totAmnt, t.unitId FROM (SELECT DISTINCT landlord_payment.id,landlord_payment.amount,landlord_payment.landlordId FROM landlord_payment) lp,

我正在尝试选择一些值,我使用了以下查询:

SELECT 
    l.*,SUM(lp.amount) AS landPayMonth,p.pName,u.uName,SUM(ci.amount) AS totAmnt,
    t.unitId
FROM
    (SELECT DISTINCT landlord_payment.id,landlord_payment.amount,landlord_payment.landlordId FROM landlord_payment) lp,
    (SELECT DISTINCT cashIn.id,cashIn.tenantId,cashIn.registeredTime,cashIn.amount FROM cashIn) ci,landlords l,properties p,units u,tenants t
WHERE
    ci.tenantId = t.id
        AND l.id = lp.landlordId
        AND t.unitId = u.id
        AND u.propertyId = p.id
        AND p.landlordId = l.id
        AND STR_TO_DATE(ci.registeredTime, '%Y-%m') = STR_TO_DATE(CURDATE(), '%Y-%m')
GROUP BY l.id;
运行上述查询后得到的部分结果是:

id  fName  lName     landPayMonth pName           uName   totAmnt   unitId  
1   Kigen  Chemweno  100000       Smart Apartments Hse 21 32200     1
landPayMonth和totAmnt产生错误的结果。其他一切都好。 部分表格的某些部分如下: 兑现

地主

id  fName   lName
1   Kigen   Chemweno
3   Samwel  Ekiru
性质

id  pName             landlordId
1   Smart Apartments    1
2   Iten Star           3
租户

id  fName       lName    unitId
1   Florence    Cheptum     1
2   Rose        Boit        2
单位

业主付款

id  landlordId  amount
1   1           20000
3   1           5000

我不知道我做错了什么

请用文字写下您希望通过此查询找到什么?我想从现金和房东付款表中获取金额总和
id  uName   propertyId
1   Hse 21  1
2   Hse 20  1
id  landlordId  amount
1   1           20000
3   1           5000