Question

我的道歉 - 我意识到这是一个非常常见的问题，但是我在查看需要使用的查询时遇到了一些麻烦。

我基本上是在尝试将值从一个表复制到另一个表，基于查询第二个表。

例如，我有两个表：

+-----------------------------------------------------+
|                       Table 1                       |
+----------+------+----------+------------+-----------+
| UniqueID | name | location | Flavour    | Status    |
| 723948   |      |          | Mango      | Tried     |
| 723948   |      |          | Orange     | Not tried |
| 723948   |      |          | Strawberry | Tried     |
| 2346     |      |          | Mango      | Not tried |
| 2346     |      |          | Strawberry | Tried     |
| 3745     |      |          | Strawberry | Tried     |
| 3745     |      |          | Mango      | Tried     |
+----------+------+----------+------------+-----------+

和

+-------------------------------------------+
|                  Table 2                  |
+----------+-----------------+--------------+
| UniqueID | fullname        | baselocation |
| 723948   | Steve Stevenson | London       |
| 2346     | Mary Marington  | New York     |
+----------+-----------------+--------------+

然后我希望根据与表2匹配的UniqueID来填充表1中的名称和位置列，因此结果如下：

+--------------------------------------------------------------------------------+
|                              Desired result (Table 1)                          |
+--------------------------+-----------------+----------+------------+-----------+
| UniqueID                 | name            | location | Flavour    | Status    |
| 723948                   | Steve Stevenson | London   | Mango      | Tried     |
| 723948                   | Steve Stevenson | London   | Orange     | Not tried |
| 723948                   | Steve Stevenson | London   | Strawberry | Tried     |
| 2346                     | Mary Marington  | New York | Mango      | Not tried |
| 2346                     | Mary Marington  | New York | Strawberry | Tried     |
| 3745                     |                 |          | Strawberry | Tried     |
| 3745                     |                 |          | Mango      | Tried     |
+--------------------------+-----------------+----------+------------+-----------+

我知道我应该使用UPDATE，SET，FROM然后加入的内容，但我不能100％确定正确，最准确/最有效的查询是什么。

谢谢！

Answer 1

您正在使用MySQL's multi-table UPDATE syntax寻找带有相当传统const paths = { src: './src', publicSrc: './public/js', dest: './app', bundle: 'bundle.js', bundleDest: './app/public/js', publicEntries: [ './public/js/index', './public/js/components/test' <--- Remove this line. ] };的{{1}}语句。

基本格式为：

UPDATE

在你的情况下：

INNER JOIN

只会修改匹配的行，这就是为什么您可以使用UPDATE T1 JOIN T2 ON T1.col = T2.col JOIN T3 ON T2.other = T3.other ... SET T1.updatedcol = T2.colvalue, T1.updatedcol2 = T2.colvalue2,执行此操作而不是UPDATE Table1 INNER JOIN Table2 ON Table1.UniqueID = Table2.UniqueID SET -- Set values in Table1 from joined rows in Table2 Table1.name = Table2.fullname, Table1.location = Table2.baselocation。

（Here is a demonstration，虽然SQLfiddle目前无法运行INNER JOIN）

Answer 2

尝试使用以下查询

UPDATE `table 1` AS t1
INNER JOIN `table 2` AS t2
ON t1.UniqueID = t2.UniqueID
SET
  name = fullname,
  location=baselocation

（我的）SQL - 根据表B填充表A中的值

2 个答案: