Question

SELECT 
      [Represantative] 
      ,[log_date] 
      ,[customer] 
      ,[type] 
      ,[log_type] 
      ,[zone_code] 
  FROM [dbo].[ilc_ziyaret_plani_gun_degisiklikleri_v] gun_logu
order by Represantative, log_date, customer
for xml path ,  root('log_details')

我需要在order by子句中嵌套的xml：我尝试使用for xml path子句，但它没有进行嵌套，它只是塑造了xml doc。 Represantative，log_date，customer

<log_details>
 <represantative name="john doe">
  <log_date date="2010-02-10">
   <customer c="xyz">
    <op>
      <log_type="add"/>
      <zone_code="a1"/>
    </op>
    <op>
      <log_type="add"/>
      <zone_code="b1"/>
    </op>
   <customer/>
   <customer> same as above </customer>
  <log_date>same for next date</log_date>

任何想法如何做到这一点

Answer 1

在TSQL中格式化xml需要一些试验和错误

SELECT  
      [Represantative] 
      ,[log_date]  
      ,[customer]  "customer/@name"
      ,[type]  "customer/@type"
      ,[log_type] "customer/op/log_type" 
      ,[zone_code]  "customer/op/zone_code"
  from t1
order by Represantative, log_date, customer 
for xml path('log_detail') ,  root('log_details')

返回此xml输出

<log_details>
  <log_detail>
    <Represantative>john doe</Represantative>
    <log_date>2010-02-10</log_date>
    <customer name="xyz" type="add">
      <op>
        <log_type>a1</log_type>
        <zone_code>b1</zone_code>
      </op>
    </customer>
  </log_detail>
</log_details>

select for xml，按字段顺序嵌套

1 个答案: