Question

我试图用另一个表中的数据更新表（使用PostgreSQL）。我的表是这样的：

+-------------------+-------------------+-----------------------+
|   id_location     |   user_location   |   social_sec_number   |
+-------------------+-------------------+-----------------------+
|   00000000001     |   Jason (null)    |   812.539.037         |
+-------------------+-------------------+-----------------------+
|   00000000002     |   Jennifer (null) |   066.307.382         |
+-------------------+-------------------+-----------------------+
|   00000000003     |   Albert (null)   |   560.732.535         |
+-------------------+-------------------+-----------------------+

我想进入这个：

+-------------------+---------------------------+-----------------------+
|   id_location     |   user_location           |   social_sec_number   |
+-------------------+---------------------------+-----------------------+
|   00000000001     |   Jason (812.539.037)     |   812.539.037         |
+-------------------+---------------------------+-----------------------+
|   00000000002     |   Jennifer (066.307.382)  |   066.307.382         |
+-------------------+---------------------------+-----------------------+
|   00000000003     |   Albert (560.732.535)    |   560.732.535         |
+-------------------+---------------------------+-----------------------+

列id_location和user_location位于同一个表TableLocation中，但social_sec_number位于另一个表中。

我的代码尝试更新它们（此代码不反映表格示例中显示的内容）：

WITH tb_cpf
AS (
    SELECT doc.nr_documento_identificacao
    FROM core.tb_usuario_localizacao ul
    INNER JOIN core.tb_localizacao l ON ul.id_localizacao = l.id_localizacao
    INNER JOIN client.tb_pess_doc_identificacao doc ON ul.id_usuario = doc.id_pessoa
    WHERE l.ds_localizacao ilike '%(null)%'
        AND doc.cd_tp_documento_identificacao = 'CPF'
    )
UPDATE core.tb_localizacao AS l
SET l.ds_localizacao = REPLACE(l.ds_localizacao, '(null)', tb_cpf)
WHERE l.id_localizacao IN (
        SELECT l.id_localizacao
        FROM core.tb_usuario_localizacao ul
        INNER JOIN core.tb_localizacao l ON ul.id_localizacao = l.id_localizacao
        INNER JOIN client.tb_pess_doc_identificacao doc ON ul.id_usuario = doc.id_pessoa
        WHERE l.ds_localizacao ilike '%(null)%'
            AND doc.cd_tp_documento_identificacao = 'CPF'
        )
    AND tb_pess_doc_identificacao.id_pessoa = tb_usuario_localizacao.id_usuario
    AND tb_usuario_localizacao.id_localizacao = tb_localizacao.id_localizacao;

我收到了这个丑陋的错误：

ERROR: column "tb_cpf" does not exist
LINE 11: ... of l.ds_localizacao = REPLACE (l.ds_localization, '(null)', tb_cpf) 
                                                                            ^
********** Error **********
ERROR: column "tb_cpf" does not exist
SQL state: 42703
Character: 433

这样，每条记录都可能带有＆＃34; null＆＃34;在每张桌子上被社会保障（例如）取代？

Answer 1

CTE中的protected void gvCalloutTeam_SelectedIndexChanged(object sender, EventArgs e) { //////////////////my logic }与主tb_cpf语句中的任何其他关系一样。您缺少列标识符，因此如果您将UPDATE添加到.nr_documento_identificacao，则更新应该有效：

tb_cpf

Answer 2

您误解了with条款的使用。可以把它想象成一个视图或像子查询一样。它的工作方式与内存表完全相同。您必须公开您将要使用的所有列。

但是，为什么这么复杂呢？为什么不做这么简单的事情？

update mytable m1

set m1.mycolumn = replace(  m1.mycolumn, 
                            '(null)', 
                            (
                            select mycolumn2 
                            from mytable2 m2 
                            where m2.id = m1.id
                            ) 
                        )

where m1.mycolumn like '%(null)'

或者，如果您真的想使用with（针对性能问题）：

with t1 as
(select t.id,
        t.mycolumn
   from mytable
)

update mytable2 t2
   set t2.mycolumn = replace (t2.mycolumn, '(null)', t1.mycolumn)
 where t2.id = t1.id
   and t2.mycolumn like '%(null)'

使用另一个表的值更新表（PostgreSQL）

2 个答案: