Question

我需要以XML格式获取表的架构。

考虑具有列的表Person：

ID varchar（10）Identity，Name varchar（20），Designation varchar（10）

我需要XML格式为

<Persons>
<Person LocalizationSetting = {today's date}>
<ID type=varchar length =10 required=true/>
<Nametype=varchar length =20 required=true/>
<Designation  type=varchar length =10 required=true/>
</Person>
</Persons>

我有以下脚本

 DECLARE @TableSchema XML

SELECT @TableSchema = (
    select  column_name,
            data_type as [type],
            CHARACTER_MAXIMUM_LENGTH AS maxLength,
            case(is_nullable)
                when 'YES' then 'false'
                else 'true'    
            end as [required]
    from information_schema.columns [column]
    where table_name = 'Person' 
    for xml auto,root('Persons') 
    )

SELECT @TableSchema

我得到以下结果：

<Patients>
  <column column_name="ID" type="varchar"  maxLength="10" required="true" />
  <column column_name="Name" type="varchar" maxLength="20" required="true" />
  <column column_name="Designation" type="varchar" maxLength="10" required="true" />
</Patients>

是否可以将column_name作为标记，如预期结果所示？

Answer 1

你应该解决这个问题......

我需要以XML格式获取表的架构。

...使用标准的内置功能。您提供的预期输出的架构可能是一些自定义的结构......请查看：

CREATE TABLE Person  (ID VARCHAR(10) NOT NULL
                     ,NameType VARCHAR(20) NOT NULL
                     ,Designation VARCHAR(10) NOT NULL);
INSERT INTO Person VALUES('SomeID','SomeType','SomeDes');

SELECT * FROM Person FOR XML AUTO,XMLDATA;

- XMLDATA返回此架构

<Schema name="Schema1" xmlns="urn:schemas-microsoft-com:xml-data" xmlns:dt="urn:schemas-microsoft-com:datatypes">
  <ElementType name="Person" content="empty" model="closed">
    <AttributeType name="ID" dt:type="string" />
    <AttributeType name="NameType" dt:type="string" />
    <AttributeType name="Designation" dt:type="string" />
    <attribute type="ID" />
    <attribute type="NameType" />
    <attribute type="Designation" />
  </ElementType>
</Schema>

- 下一个可能性

SELECT * FROM Person FOR XML AUTO,XMLSCHEMA;

- XMLSCHEMA返回此架构

<xsd:schema targetNamespace="urn:schemas-microsoft-com:sql:SqlRowSet1" xmlns:schema="urn:schemas-microsoft-com:sql:SqlRowSet1" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:sqltypes="http://schemas.microsoft.com/sqlserver/2004/sqltypes" elementFormDefault="qualified">
  <xsd:import namespace="http://schemas.microsoft.com/sqlserver/2004/sqltypes" schemaLocation="http://schemas.microsoft.com/sqlserver/2004/sqltypes/sqltypes.xsd" />
  <xsd:element name="Person">
    <xsd:complexType>
      <xsd:attribute name="ID" use="required">
        <xsd:simpleType>
          <xsd:restriction base="sqltypes:varchar" sqltypes:localeId="1033" sqltypes:sqlCompareOptions="IgnoreCase IgnoreKanaType IgnoreWidth">
            <xsd:maxLength value="10" />
          </xsd:restriction>
        </xsd:simpleType>
      </xsd:attribute>
      <xsd:attribute name="NameType" use="required">
        <xsd:simpleType>
          <xsd:restriction base="sqltypes:varchar" sqltypes:localeId="1033" sqltypes:sqlCompareOptions="IgnoreCase IgnoreKanaType IgnoreWidth">
            <xsd:maxLength value="20" />
          </xsd:restriction>
        </xsd:simpleType>
      </xsd:attribute>
      <xsd:attribute name="Designation" use="required">
        <xsd:simpleType>
          <xsd:restriction base="sqltypes:varchar" sqltypes:localeId="1033" sqltypes:sqlCompareOptions="IgnoreCase IgnoreKanaType IgnoreWidth">
            <xsd:maxLength value="10" />
          </xsd:restriction>
        </xsd:simpleType>
      </xsd:attribute>
    </xsd:complexType>
  </xsd:element>
</xsd:schema>

如果您确实需要填写预期的格式，那么本机查询不支持动态列名。作为一种解决方法，您可以使用动态创建的SQL与EXEC或此方法一起运行，这种方法在字符串级别上使用非常难看的黑客攻击（最后应该避免使用XML！）：

WITH AlmostCorrect AS
(
SELECT
    (
    SELECT  GETDATE() AS [@LocalizationSetting]
           ,(
            SELECT   COLUMN_NAME AS [@name]
                    ,DATA_TYPE AS [@type]
                    ,CHARACTER_MAXIMUM_LENGTH AS [@length]
                    ,CASE IS_NULLABLE
                        WHEN 'YES' THEN 'false'
                        ELSE 'true'    
                     END AS [@required]
            FROM INFORMATION_SCHEMA.COLUMNS
            WHERE TABLE_NAME = 'Person' 
            FOR XML PATH('row'),TYPE
            ) AS [*]
    FOR XML PATH('Person'),ROOT('Persons'),TYPE
    ) AS TheXML
)
SELECT CAST(REPLACE(REPLACE(CAST(TheXML AS NVARCHAR(MAX)),'<row name="','<'),'" type',' type') AS XML)
FROM AlmostCorrect

结果

<Persons>
  <Person LocalizationSetting="2017-04-18T11:05:30.047">
    <ID type="varchar" length="10" required="true" />
    <NameType type="varchar" length="20" required="true" />
    <Designation type="varchar" length="10" required="true" />
  </Person>
</Persons>

使用表列名称作为XML标记在SQL中构造XML

1 个答案: