Question

我想知道它是否足够向前以以下方式创建视图，我的尝试没有按预期进行，

我正在尝试使用以下查询创建另一个视图，以在满足以下条件时标识有问题的“ grp”：

我对病情的想法的原始描述：

从控件WHERE中选择grp

（'control_1.pvalue'||'（2nd）'！='control_2.pvalue'）

OR

（'control_1.pvalue'||'（3rd）'！='control_3.pvalue'）

OR

（'control_1.pvalue'||'（4th）'！='control_4.pvalue'）

OR

（'control_1.pvalue'||'（5th）'！='control_5.pvalue'）

OR

（'control_1.pvalue'||'（6th）'！='control_6.pvalue'）

我尝试了以下操作，希望返回 M2 ：

SELECT grp FROM control_1 WHERE (pvalue || " (2nd)" != (SELECT pvalue from control_2) AND grp = (SELECT grp from control_2));

但是，即使这个基本步骤对我也不起作用。

运行下面的代码片段以查看我要尝试使用的表和视图。

th, td {border: 1px solid black;}

<div>
<div>CREATE TABLE control (</div>
<div>id INTEGER PRIMARY KEY AUTOINCREMENT,</div>
<div>grp TEXT,</div>
<div>pname TEXT,</div>
<div>pvalue TEXT);</div>
<div></div><br />

<table>
<tr><td colspan="4"><div>control</div></td></tr>
<tr><th>id</th><th>grp</th><th>pname</th><th>pvalue</th></tr>
<tr><td>1</td><td>M1</td><td>d1</td><td>vat</td></tr>
<tr><td>2</td><td>M1</td><td>d2</td><td>vat (2nd)</td></tr>
<tr><td>3</td><td>M1</td><td>d3</td><td>vat (3rd)</td></tr>
<tr><td>4</td><td>M1</td><td>d4</td><td>vat (4th)</td></tr>
<tr><td>5</td><td>M1</td><td>d5</td><td>vat (5th)</td></tr>
<tr><td>6</td><td>M1</td><td>d6</td><td>vat (6th)</td></tr>
<tr><td>7</td><td>M2</td><td>d1</td><td>bin</td></tr>
<tr><td>8</td><td>M2</td><td>d2</td><td>ban (2nd)</td></tr>
<tr><td>9</td><td>M2</td><td>d3</td><td>bin (3rd)</td></tr>
<tr><td>10</td><td>M2</td><td>d4</td><td>bin (4th)</td></tr>
<tr><td>11</td><td>M2</td><td>d5</td><td>bin (5th)</td></tr>
<tr><td>12</td><td>M2</td><td>d6</td><td>bin (6th)</td></tr>
<tr><td>13</td><td>M3</td><td>d1</td><td>haw</td></tr>
<tr><td>14</td><td>M3</td><td>d2</td><td>haw (2nd)</td></tr>
<tr><td>15</td><td>M3</td><td>d3</td><td>ha (3rd)</td></tr>
<tr><td>16</td><td>M3</td><td>d4</td><td>haw (4th)</td></tr>
<tr><td>17</td><td>M3</td><td>d5</td><td>haw (5th)</td></tr>
<tr><td>18</td><td>M3</td><td>d6</td><td>haw (6th)</td></tr>
</table>
<div></div><br />
<div>CREATE VIEW control_1 AS SELECT * FROM control WHERE pname = "d1";</div>
<div></div>

<table>
<tr><th colspan="4">control_1</th></tr>
<tr><th>id</th><th>grp</th><th>pname</th><th>pvalue</th></tr>
<tr><td>1</td><td>M1</td><td>d1</td><td>vat</td></tr>
<tr><td>7</td><td>M2</td><td>d1</td><td>bin</td></tr>
<tr><td>13</td><td>M3</td><td>d1</td><td>haw</td></tr>
</table>
<div></div><br />
<div></div>
<div>CREATE VIEW control_1 AS SELECT * FROM control WHERE pname = "d2";</div>
<div></div>

<table>
<tr><th colspan="4">control_2</th></tr>
<tr><th>id</th><th>grp</th><th>pname</th><th>pvalue</th></tr>
<tr><td>2</td><td>M1</td><td>d2</td><td>vat (2nd)</td></tr>
<tr><td>8</td><td>M2</td><td>d2</td><td>ban (2nd)</td></tr>
<tr><td>14</td><td>M3</td><td>d2</td><td>haw (2nd)</td></tr>
</table>
<div></div><br />
<div>
<div>CREATE VIEW control_1 AS SELECT * FROM control WHERE pname = "d3";</div>
<div></div>

<table>
<tr><th colspan="4">control_3</th></tr>
<tr><th>id</th><th>grp</th><th>pname</th><th>pvalue</th></tr>
<tr><td>3</td><td>M1</td><td>d3</td><td>vat (3rd)</td></tr>
<tr><td>9</td><td>M2</td><td>d3</td><td>bin (3rd)</td></tr>
<tr><td>15</td><td>M3</td><td>d3</td><td>ha (3rd)</td></tr>
</table>
<div></div><br />

<div>
<div>--I create the following as well, but we can see the problems from the 1st three views</div>
<div>CREATE VIEW control_1 AS SELECT * FROM control WHERE pname = "d4";</div>
<div>CREATE VIEW control_1 AS SELECT * FROM control WHERE pname = "d5";</div>
<div>CREATE VIEW control_1 AS SELECT * FROM control WHERE pname = "d6";</div>
</div>

Answer 1

我有一个顿悟，我能够找出以下几点：

我创建了两个新的视图，如下所示，现在我可以得到期望的结果。

.then

这是此新视图上的select语句：

create view dview as
SELECT
DISTINCT
grp,
(select pvalue from control_1 WHERE control.grp == grp) as d1,
(select pvalue from control_2 WHERE control.grp == grp) as d2,
(select pvalue from control_3 WHERE control.grp == grp) as d3,
(select pvalue from control_4 WHERE control.grp == grp) as d4,
(select pvalue from control_5 WHERE control.grp == grp) as d5,
(select pvalue from control_6 WHERE control.grp == grp) as d6
FROM control;

开始创建新视图：

select * from dview;
grp         d1          d2          d3          d4          d5          d6
----------  ----------  ----------  ----------  ----------  ----------  ----------
M1          vat         vat (2nd)   vat (3rd)   vat (4th)   vat (5th)   vat (6th)
M2          bin         ban (2nd)   bin (3rd)   bin (4th)   bin (5th)   bin (6th)
M3          haw         haw (2nd)   ha (3rd)    haw (4th)   haw (5th)   haw (6th)

这是此新视图上的select语句：

create view dview_err as
SELECT * FROM dview WHERE
(d1 || " (2nd)" != d2) OR
(d1 || " (3rd)" != d3) OR
(d1 || " (4th)" != d4) OR
(d1 || " (5th)" != d5) OR
(d1 || " (6th)" != d6);

尝试在SQLite3中创建具有多个条件的SQL VIEW

1 个答案: