如果找到不存在的键，则使用已知键更新多行而不插入新行

2022-01-09 00:00:00 insert mysql

假设我们有表格items...

table: items
item_id INT PRIMARY AUTO_INCREMENT
title VARCHAR(255)
views INT

让我们想象它充满了类似的东西

Let's imagine that it is filled with something like

(1, item-1, 10),
(2, item-2, 10),
(3, item-3, 15)

我想根据从这个数组 [item_id] => [views] 获取的数据为这些项目创建 multi 更新视图

I want to make multi update view for this items from data taken from this array [item_id] => [views]

'1' => '50',
'2' => '60',
'3' => '70',
'5' => '10'

重要！请注意，我们在数组中有 item_id=5，但在数据库中没有 item_id=5.

IMPORTANT! Please note that we have item_id=5 in array, but we don't have item_id=5 in database.

我可以使用 INSERT ... ON DUPLICATE KEY UPDATE，但这样 image_id=5 将被插入到 talbe 项中.如何避免插入新密钥?我只想跳过 item_id=5，因为它不在表中.

I can use INSERT ... ON DUPLICATE KEY UPDATE, but this way image_id=5 will be inserted into talbe items. How to avoid inserting new key? I just want item_id=5 be skipped because it is not in table.

当然，在执行之前，我可以从项目表中选择现有的键；然后与数组中的键进行比较；删除不存在的密钥并执行 INSERT ... ON DUPLICATE KEY UPDATE.但也许有一些更优雅的解决方案?

Of course, before execution I can select existing keys from items table; then compare with keys in array; delete nonexistent keys and perform INSERT ... ON DUPLICATE KEY UPDATE. But maybe there is some more elegant solutions?

谢谢.

推荐答案

您可以尝试生成一个文字表并通过加入该表来更新项目:

You may try to generate a table of literals and update items by joining with the table:

UPDATE items
    JOIN (SELECT 1 as item_id, 50 as views
          UNION ALL
          SELECT 2 as item_id, 60 as views
          UNION ALL
          SELECT 3 as item_id, 70 as views
          UNION ALL
          SELECT 5 as item_id, 10 as views
          ) as updates
         USING(item_id)
 SET items.views = updates.views;

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