Question

我有这个SQL表：

GAME

ID     PLAYER     MIN_VALUE_1  MIN_VALUE_2
1      John       1.2          3
2      Paul       1.4          3.2
3      Henry      1.6          3.4
4      Tom        2.1          3.6

PREDICTIONS

ID_GAME     HOUSE       VALUE_1   VALUE_2    
1           house_a     1.5       3
1           house_b     1         2.9
2           house_a     1.2       3.1
2           house_b     1.4       3.6
2           house_c     1.5       4
3           house_a     1.5       4
3           house_b     1.7       4.2
3           house_c     1.8       4
3           house_d     1.9       4.3
4           house_b     1.6       3.1
4           house_d     1.9       3

我有两个选择：

有了这个：

SELECT *                  
FROM game m
INNER JOIN predictions p ON m.id = p.id_game
WHERE p.value_1 >= m.min_value_1 OR p.value_2 >= m.min_value_2

我获取了VALUE_1＆gt; = MIN_VALUE_1或VALUE_2＆gt; = MIN_VALUE_2

的所有行

ID_GAME     HOUSE       VALUE_1   VALUE_2    MIN_VALUE_1   MIN_VALUE_2
1           house_a     1.5       3          1.2          3
2           house_b     1.4       3.6        1.4          3.2
2           house_c     1.5       4          1.4          3.2
3           house_a     1.5       4          1.6          3.4
3           house_b     1.7       4.2        1.6          3.4
3           house_c     1.8       4          1.6          3.4
3           house_d     1.9       4.3        1.6          3.4

并查询：

SELECT *                  
FROM game m
INNER JOIN predictions p ON m.id = p.id_game
WHERE p.value_1 < m.min_value_1 AND p.value_2 < m.min_value_2

我获得VALUE_1＆lt;所有行MIN_VALUE_1和VALUE_2＆lt; MIN_VALUE_2

ID_GAME     HOUSE       VALUE_1   VALUE_2    MIN_VALUE_1   MIN_VALUE_2

1           house_b     1         2.9        1.2          3
2           house_a     1.2       3.1        1.4          3.2    
4           house_b     1.6       3.1        2.1          3.6
4           house_d     1.9       3          2.1          3.6

但我需要获取两个查询的所有行，但相同的ID_GAME 。我需要这个：

ID_GAME     HOUSE       VALUE_1   VALUE_2    MIN_VALUE_1   MIN_VALUE_2
1           house_a     1.5       3          1.2          3
1           house_b     1         2.9        1.2          3
2           house_a     1.2       3.1        1.4          3.2    
2           house_b     1.4       3.6        1.4          3.2
2           house_c     1.5       4          1.4          3.2
3           house_a     1.5       4          1.6          3.4
3           house_b     1.7       4.2        1.6          3.4
3           house_c     1.8       4          1.6          3.4
3           house_d     1.9       4.3        1.6          3.4

Answer 1

我认为你应该删除where条款

select * from predictions p INNER JOIN game m ON p.id_game = m.id WHERE  
p.value_1 < m.min_value_1 AND p.value_2 < m.min_value_2;

Answer 2

最后我找到了一个解决方案（我不知道是否理想的解决方案）：

ngOnChanges(changes: any): void {
    console.log('change', changes.skills.currentValue);
}

MySQL加入两个具有相同ID的选择

2 个答案: