Question

我需要表T1和T2

T1

Id name open   receive close
1  jon  5000    4000   1000
2  jim  6000    5000   1000

T2

Id name receive
1  jon   500
2  jim   700

我想要这样（T1接收+ T2接收）结果

Id name  open   receive close
1  jon   5000   4500    500
2  jim   6000   5700    300

我正在使用Sqlite

Answer 1

更新：Tim Biegeleisen的回答更为正确。

你需要在这两个表上进行连接：

SELECT T1.id, T1.open, T1.receive+T2.receive, T1.close
FROM T1 JOIN T2 USING (id);

修改：如果您想要更新T1，可以这样做：

UPDATE T1
SET receive=receive+coalesce((SELECT receive FROM T2 WHERE T1.id=T2.id),0);

Answer 2

SELECT T1.name,
       T1.open,
       T1.receive + COALESCE(T2.receive, 0) AS receive,
       T1.close - COALESCE(T2.receive, 0)   AS close
FROM T1
LEFT JOIN T2
    ON T1.Id = T2.Id

如果你想UPDATE第一张表，你可以试试这个：

UPDATE T1
SET T1.recieve = T1.recieve +
    (SELECT COALESCE(T2.recieve, 0) FROM T2 WHERE T2.Id = T1.Id),
    T1.close = T1.close -
    (SELECT COALESCE(T2.recieve, 0) FROM T2 WHERE T2.Id = T1.Id)

Answer 3

如果每id只有一行，则可以执行以下操作：

select t1.id, t1.name, t1.open, (t1.receive + t2.receive), t1.close
from t1 join
     t2
     on t1.id = t2.id;

我可以推测t2实际上对于给定的id或缺少的行有多行。这表明：

select t1.id, t1.name, t1.open, (t1.receive + t2.receive), t1.close
from t1 left join
     (select t2.id, sum(t2.receive) as receive
      from t2
      group by t2.id
     ) t2
     on t1.id = t2.id;

如何将一个列值与其他列值相加

3 个答案: