根据另一个表中的数据更新 SQLite 表 [英] Update SQLite table based on data in another table

问题描述

我在 SQLite 中有两个表，看起来像这样

I have two tables in SQLite which look like this

TABLE_X    
____________________
| id | C1 | C2 | C3 | C4 |
| 10 | 99 | 03 | 04 | 00 |
| 60 | 88 | 20 | 30 | 40 |


TABLE_Y
___________     
| id | C2 |
| 10 | 11 |
| 60 | 22 |

我正在尝试编写查询以根据表 Y 中的记录更新表 X 上的记录更新条件如下

I am trying to write query to update records on Table X based on records in Table Y The condition for the updateis something like the following

update table_x
   set table_x.c1 = 100,
       table_x.c2 = table_y.c2
 where table_x.id = table_y.id

但是当我尝试这样做时，我收到一条错误消息

But when I try to do this I get an error message saying

没有这样的列:table_y.c2

推荐答案

已删除的答案关于错误原因是正确的:关系标识符必须引入(例如使用 FROM/JOIN)可以使用之前的查询.

The deleted answer was correct about the cause of the error: a relation identifier must be introduced (e.g. with FROM/JOIN) in a query before it can be used.

虽然SQLite不支持UPDATE..JOIN(所以没有办法直接引入查找关系)，可以使用依赖子查询来模拟效果:

While SQLite does not support UPDATE..JOIN (so there is no way to introduce the lookup relation directly), a dependent sub-query can be used to simulate the effect:

update table_x
   set c1 = 100,
       c2 = (select y.c2 from table_y as y
             where y.id = table_x.id)

注意，不像一个正确的UPDATE..JOIN，如果子选择未能找到匹配项，则将分配NULL.

Note that unlike a proper UPDATE..JOIN, if the sub-select fails to find a match then NULL will be assigned.

天啊.

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