XML 存在方法不过滤 SQL Server 2012 中的记录

Question

我有两个问题：

1. 我正在尝试在下面的示例中过滤属性@Type = "Bikes" 。 但我得到了所有 6 行。 我应该只得到 3 行。 为什么不在这里工作，我不知道？

2. 如何进一步过滤@Type = "Bikes" and Product = "Mountain"

我的代码如下：

declare @xml xml = '<StoreSurvey>
    <AnnualSales>800000</AnnualSales>
    <AnnualRevenue>80000</AnnualRevenue>
    <BankName>United Security</BankName>
    <BusinessType>BM</BusinessType>
    <YearOpened>1996</YearOpened>
    <Specialty>Mountain</Specialty>
    <SquareFeet>21000</SquareFeet>
    <Brands>2</Brands>
    <Internet>ISDN</Internet>
    <NumberEmployees>13</NumberEmployees>
    <Products Type="Bikes">
      <Product>Mountain</Product>
      <Product>Road</Product>
      <Product>Racing</Product>
    </Products>
    <Products Type="Clothes">
      <Product>Jerseys</Product>
      <Product>Jackets</Product>
      <Product>Shorts</Product>
    </Products>
  </StoreSurvey>'    

问题 1 的代码：

  select T.col.value('.','varchar(100)')  
  from @xml.nodes('/StoreSurvey/Products/Product') as T(col)
  where T.col.exist('/StoreSurvey/Products[@Type = "Bikes"]') =1

期望的输出：

Mountain
Road
Racing

问题 2 的代码：此代码导致“语法错误”：

  select T.col.value('.','varchar(100)')  
  from @xml.nodes('/StoreSurvey/Products/Product') as T(col)
  where T.col.exist('/StoreSurvey/Products[@Type = "Bikes"]/[Product = "Mountain"]') = 1

Answer 1

这两个都返回您想要的第一个结果

  select T.col.value('.','varchar(100)')  
  from @xml.nodes('/StoreSurvey/Products/Product') as T(col)
  where T.col.exist('..[@Type = "Bikes"]') =1

第二个看起来更自然，因为获取所有Product节点然后备份并在父轴上应用过滤器很奇怪（使用.. ）

  select T.col.value('.','varchar(100)')  
  from @xml.nodes('/StoreSurvey/Products[@Type = "Bikes"]/Product') as T(col)

目前尚不清楚您对 Q2 的确切期望结果是什么，但这会返回一个结果

select T.col.value('.','varchar(100)')  
from @xml.nodes('/StoreSurvey/Products[@Type = "Bikes"]/Product[text() = "Mountain"]') as T(col)

Answer 2

对于这两种情况，请使用如下查询表单：

with q as
(
  select products.col.value('@Type','varchar(100)')  ProductType, 
         product.col.value('.','varchar(100)') Product
  from @xml.nodes('/StoreSurvey/Products') as products(col)
  cross apply products.col.nodes('Product') as product(col)
)
select * 
from q
  where ProductType = 'Bikes'
    and Product = 'Mountain';