我有两种关系观点。
第一种观点：
CREATE VIEW VIEW1
AS
SELECT T#,
MAX(DECODE(LEG#,1,DEPARTURE)) ORIGIN,
MAX(DECODE(LEG#,1,DESTINATION)) DESTINATION1
FROM TRIPLEG
WHERE T# IN
(SELECT T# FROM TRIPLEG WHERE LEG# < 3
AND T# IN
(SELECT T# FROM TRIPLEG GROUP BY T#
HAVING COUNT(T#) < 3)
GROUP BY T#)
GROUP BY T#
ORDER BY T#;
第二种观点：
CREATE VIEW VIEW2
AS
SELECT T#,
MAX(DECODE(LEG#,2,DESTINATION)) DESTINATION1
FROM TRIPLEG
WHERE T# IN
(SELECT T# FROM TRIPLEG WHERE LEG# < 3
AND T# IN
(SELECT T# FROM TRIPLEG GROUP BY T# …我有两个:
bool isPointOnShape(int a, int b)
{
}
bool isPointInShape(int a, int b)
{
}
假设我有一个正方形,第一个点(左下角)是x,y(0,0)第二个点(左上角)是(0,2),第三个是(2,2),第四个是(0,2) .
形状上的点将是(0,1)(1,2)(2,1)(1,0)并且形状中的点是(1,1)
如何找出形状/形状上的点并返回真值,以便我可以将它存储在某个地方？