如何从父表的两个子表中获取值

时间:2018-12-24 08:04:32

标签: mysql

我有三个表格(发票,费率,费率拍打) 我在发票表中有重量和no_of_package 如果invoice.weight <= rate.weight,则它将返回该表rate.weight 否则weight <= rate_slap.weight然后它将返回该表rate_slap.weight

但不从两个表中返回任何一个表值。

SELECT 
rate.`minimum_wgt`, 
slap.`slap_weight`
FROM m_ac_customer_rate rate
LEFT JOIN m_ac_customer_rate_slap slap ON (rate.minimum_wgt<=15) 
OR (slap.slap_weight<=15)
WHERE rate.customer_rate_id=slap.customer_rate_master 
AND `customer_name`=1007
AND `destination_name`=3
AND `service_type`=1
AND `shipment_type`='D'
AND `payment_mode`='CASH';

样本数据

发票表

invoice_no,weight,total_carton
 2142423,   10.4,     5

费率表

invoice_no,weight,rate_per_kg
2142423 , 15.8,      150.00

rate_slap表

rateslap_id,weight,rate_per_kg
2142423 , 10.8,      10.00

1 个答案:

答案 0 :(得分:0)

这将在子查询中选择两个权重(如果存在)以及发票权重,然后在主查询中执行CASE ...以选择正确的权重

SELECT no_of_package, 
  CASE WHEN weight <= rate_weight THEN rate_weight
       WHEN weight <= slap_weight THEN slap_weight
       ELSE 0
       END AS other_weight
FROM 
  (SELECT i.weight, IFNULL(r.weight, 0) as rate_weight, IFNULL(rs.weight, 0) as slap_weight
   FROM invoice I
   LEFT JOIN rate r ON r.invoice_no = i.invoice_no AND minimum_wgt <= 15
   LEFT JOIN rate_slap rs ON rs.invoice_no = i.invoice_no AND slap_weight <= 15
   WHERE customer_name = 1007
     AND destination_name = 3
     AND service_type = 1
     AND shipment_type = 'D'
     AND payment_mode = 'CASH')

我为CASE选择了默认值0,也许应该是发票的重量。您需要调整表名称,我使用了名称的简短版本