本文介绍了在两个表中做一些值的计算,并将其存储在第三个表mysql中的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
table1
| email | result | ---------------------------- | abc@gmail |0.12 | | dsv@gmail |0.23 | | rthgmail | 0.45 | | hfg@gmail |0.56 | | yyt@gmail | 0.78 | | hjg@gmail | 0.35 |table2
| resource | ------------------- | 0.5 |table3
| temp | ------------------- |NULL |现在我想分别从table1和table2中为email abc@gmail添加结果和资源将结果保存在table3中。我想每次都更新table3。
Now I want addition of result and resource from table1 and table2 respectively for email abc@gmail and save the result in table3. I want to update the table3 every time.
我试过下面但是不工作:
I tried the following but its not working:
UPDATE table3 SET temp = table1.result + table2.resource WHERE email = 'abc@gmail'$ b $
推荐答案您可以使用子查询:
UPDATE Table3 t SET t.temp = (SELECT s.result+p.resource FROM table1 s INNER JOIN table2 p ON(s.email = 'abc@gmail'))如果您的Table3没有数据:
If your Table3 doesn't have data yet:
INSERT INTO Table3 (SELECT s.result+p.resource FROM table1 s INNER JOIN table2 p ON(s.email = 'abc@gmail'))更多推荐
在两个表中做一些值的计算,并将其存储在第三个表mysql中
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