使用UNION ALL为来自不同表的ID添加自动递增列

时间:2019-01-09 20:35:28

标签: php mysql jstree

我已经看到SQL Server中有一个名为 ROW_NUMBER 的函数,在我的情况下,我要联接3个表,每个表的ID列都有不同的名称(为 jsTree plugin

我有此代码:

$sql= "(SELECT
                c1.idcarrera AS id,
                c1.nombre as text,
                null as parent_id
        FROM siiupv.carrera AS c1)";
$sql.=" UNION ALL ";
$sql.= " (SELECT
                pe.idplan_estudios AS id,
                pe.clave AS text,
                pe.idcarrera as parent_id
        FROM siiupv.plan_estudios AS pe)";
$sql.=" UNION ALL ";
$sql.= " (SELECT idcarga AS id, c2.clave AS text, idplan_estudios as parent_id FROM siiupv.carga AS c2)";
$qResults = $pdo->prepare($sql);
$qResults->execute();
$count = $qResults->rowCount();
while($m = $qResults->fetch(PDO::FETCH_ASSOC)){ 
        echo '<tr><td>&nbsp;'.$m["id"].'</td><td>'.$m["text"].'</td><td>'.$m["parent_id"].'</td></tr>';
}

它产生以下输出:

+------+------------------+
| id   | text | parent_id |
+------+------------------+
|    1 |  A1  | NULL      |
|    2 |  A2  | NULL      |
|    3 |  A3  | NULL      |
|    1 |  B1  |   1       |
|    2 |  B2  |   2       |
|    3 |  B3  |   1       |
|    4 |  B4  |   2       |
|    5 |  B5  |   5       |
|    1 |  C1  |   4       | Note! (Child of "B4")
|    2 |  C2  |   4       | Note! (Child of "B4")
+------+------------------+

但是我想得到:

+------+------------------+
| id   | text | parent_id |
+------+------------------+
|    1 |  A1  | NULL      |
|    2 |  A2  | NULL      |
|    3 |  A3  | NULL      |
|    4 |  B1  |   1       |
|    5 |  B2  |   2       |
|    6 |  B3  |   1       |
|    7 |  B4  |   2       |
|    8 |  B5  |   5       |
|    9 |  C1  |   7       | Note! (Child of "B4")
|   10 |  C2  |   7       | Note! (Child of "B4")
+------+------------------+

注意!>这些ID已遍历,因此必须具有值7

我想至少获得自动增量ID的帮助

2 个答案:

答案 0 :(得分:1)

向结果集的每个记录添加行号的一种方法是使用会话变量:

"Great Britain"

答案 1 :(得分:0)

不确定我是否真的遇到问题,但是您可以通过添加前缀或其他内容来生成新的唯一ID。

选择CONCAT(“ A-”,idcarga)AS ID,c2.clave AS文本,CONCAT(“ A-”,idplan_estudios)作为parent_id