以下是样本选择1:
aStarResult=aStarSearch(tileAStarSearch, frontier);
File "C:\Users\zjalmahmoud\workspace\Tile_Problem\search.py", line 211, in aStarSearch
frontier.push(c) #add it to the queue
File "C:\Users\zjalmahmoud\workspace\Tile_Problem\utilsSimple.py", line 61, in push
insort(self.q, (self.f(item), item))
TypeError: unorderable types: Node() < Node()
这是第二个:
SELECT A, B FROM Table1
选择1和选择2的记录数量是相同的,结果应该是:
SELECT C, D FROM Table2
好的,我提供了一些数据结果:
结果1:
A, B, C, D
结果2:
+---+---+
| A | B |
+---+---+
| 1 | 2 |
| 3 | 4 |
| 5 | 6 |
+---+---+
最终结果应该是:
+---+----+
| C | D |
+---+----+
| 4 | 8 |
| 5 | 9 |
| 6 | 10 |
+---+----+
答案 0 :(得分:3)
创意:使用您可以加入的唯一密钥,例如:行号/等级。这是一个有效的例子:(SQLFiddle)
set @rank1 := 0;
set @rank2 := 0;
select a, b, c, d
from
(select @rank1 := @rank1 + 1 AS rank, a, b from table1) t1 inner join
(select @rank2 := @rank2 + 1 AS rank, c, d from table2) t2 on t2.rank = t1.rank
order by t1.rank