Question

更新表'委托'中的某些行时遇到问题。现在我有一个像bar / 123/456这样的Commission_number。我想将它重命名为John / 123/456等，从'user'表中取一个名字

commission                       user
-------------------------        -----------------
commission_number|user_id         username|user_id
-------------------------        -----------------
bar/123/456      | 1               John   | 1
bar/123/123      | 2               Bob    | 2
bar/456/123      | 3               Thomas | 3

以下是我的查询，使用'baz'。但是不知道如何从'user'表中放置'username'。我只需要替换'baz'，并使用'user'中的用户名，但不知道如何。

UPDATE commission
SET commission_number = overlay(commission_number placing 'baz' FROM POSITION('bar' IN commission_number) for 3)
WHERE commission_number in (1,2,3,4,5,6,7,8)

Answer 1

update commission 
   set commission_number = u.username || '/' || left(commission_number, -strpos(commission_number, '/'))
from users u
  where u.user_id = commission.user_id

Answer 2

我首先尝试类似......

UPDATE commission
SET commission_number = 
  (SELECT UserName FROM User WHERE user.user_id=commission.user_id)
   ||SubString(commission_number,4,8000)

还有right(commission_number,-3)除了前三个字符之外的所有字符。

我不确定你要做什么 WHERE commission_number in (1,2,3,4,5,6,7,8)？

Answer 3

这说明了更新后数据的外观。当我使用字符串函数时，我经常使用这种select语句。

select c.commission_number, c.user_id, 
       (select username 
        from "user" 
        where "user".user_id = c.user_id) username,
       overlay(c.commission_number 
               placing (select username 
                        from "user" 
                        where "user".user_id = c.user_id) 
               from 1 for 3)
from commission c;

现在，您可以按照您知道可以使用的术语编写更新语句。

update commission 
set commission_number = overlay(commission_number 
                                placing (select username 
                                         from "user" 
                                         where "user".user_id = commission.user_id) 
                                from 1 for 3);

Answer 4

尝试此查询：

 update commission
 set commission_number=user_table.user_name+
  SUBSTRING(SUBSTRING(commission.commission_number,CHARINDEX('/',commission.commission_number,1)+0,LEN(commission.commission_number))
                ,1
                ,CHARINDEX('/',SUBSTRING(commission.commission_number,CHARINDEX('/',commission.commission_number,1)+1,LEN(commission.commission_number)),1)+LEN(commission.commission_number))
 from commission
 inner join user_table on user_table.user_id=commission.user_id

在 Sql Fiddle

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花了一些时间在 SQL小提琴

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Postgres - 更改列数据

4 个答案: