如何将客户数据与联系人数据一起返回，以及如果客户没有联系人，则如何返回客户数据

问题描述

在我的应用程序中，我有两个名为clients和contacts的表，而clients可以有很多contacts，我们可以知道在contacts内部拥有client_id，我想创建一个查询到名为/clients/:id的特定路由，该查询将返回客户端数据连接以及该客户端具有的所有联系人。所以我想退货

{
    client: [
     {
      name_client:...,email_client:...,telephone_client:....,contacts: [
        allcontacts
      ]
     }
    ]
}

但是我没有得到想要的东西，我发现了与此查询相关的其他问题

select
    distinct clients.id as client_id,clients.name as client_name,clients.email as client_email,clients.telephone as client_telephone,contacts.id as contact_id,contacts.name as contact_name
from
    contacts
 inner join
    clients
    on contacts.client_id = clients.id
where
    contacts.client_id = 'd9935b6d-0497-4a45-b472-ed3b4f85cd60'

如果我的客户没有联系人，将返回一个空对象，但我不是，我想返回客户数据。

解决方法

请将您的查询更改为此：

select distinct clients.id as client_id,clients.name as client_name,clients.email as client_email,clients.telephone as client_telephone,contacts.id as contact_id,contacts.name as contact_name
  from clients
       left join contacts
              on contacts.client_id = clients.id
 where clients.client_id = 'd9935b6d-0497-4a45-b472-ed3b4f85cd60'

left join返回所有匹配的clients行（大概只有一个）以及任何contacts行具有匹配的client_id。如果没有这样的行，则contacts.id和contacts.name将是null。

如果您想让PostgreSQL为此构建一个json文档，请使用jsonb函数为您完成此操作：

with base as (
  select distinct clients.id as client_id,contacts.name as contact_name
    from clients
         left join contacts
                on contacts.client_id = clients.id
   where clients.client_id = 'd9935b6d-0497-4a45-b472-ed3b4f85cd60'
),aggregate_contacts as (
  select client_id,client_name,client_email,client_telephone,jsonb_agg(
           case 
             when contact_id is null then '{}'::jsonb
             else jsonb_build_object( 
                'contact_id',contact_id,'contact_name',contact_name
             )
           end
         ) as contacts
    from base
   group by client_id,client_telephone
)
select to_jsonb(aggregate_contacts) as result
  from aggregate_contacts;

使用lateral join：

select to_jsonb(q) as result
  from (select clients.id as client_id,case
                 when jsonb_agg(x) = '[null]'::jsonb then '[]'::jsonb
                 else jsonb_agg(x)
               end as contacts
          from clients 
               left join lateral 
                    (select id as contact_id,name as contact_name
                       from contacts
                      where id = clients.id) x on true
         where client_id = 'd9935b6d-0497-4a45-b472-ed3b4f85cd60'
         group by clients.id,clients.name,clients.email,clients.telephone
       ) as q