我想设置一个处理指令,在XML上包含一个样式表:
同样的问题是xml声明(例如this.dataGridView1.Columns.Clear();
this.dataGridView1.AutoGenerateColumns = false;
//Create Column
var column1 = new DataGridViewTextBoxColumn()
{
Name = "firstNameColumn", /*Name of Column*/
HeaderText = "First Name", /*Title of Column*/
DataPropertyName = "FirstName" /*Name of the property to bind to cilumn*/
};
//Add column to grid
this.dataGridView1.Columns.Add(column1);
)
期望的结果:
<?xml version="1.0" encoding="utf-8"?>
我的研究为我带来了节点测试语法和<?xml-stylesheet type="text/xsl" href="stylesheet.xsl"?>
<TestPath>
<Test>Test</Test>
<SomeMore>SomeMore</SomeMore>
</TestPath>
。
此
processing-instruction()
产生这个:
SELECT 'type="text/xsl" href="stylesheet.xsl"' AS [processing-instruction(xml-stylesheet)]
,'Test' AS Test
,'SomeMore' AS SomeMore
FOR XML PATH('TestPath')
我找到的所有提示告诉我将XML转换为VARCHAR,“手动”连接它并将其转换回XML。但这是 - 怎么说 - 难看?
这很明显:
<TestPath>
<?xml-stylesheet type="text/xsl" href="stylesheet.xsl"?>
<Test>Test</Test>
<SomeMore>SomeMore</SomeMore>
</TestPath>
有机会解决这个问题吗?
答案 0 :(得分:2)
还有另外一种方法,它需要两个步骤,但不需要你在过程中的任何地方将XML视为字符串:
declare @result XML =
(
SELECT
'Test' AS Test,
'SomeMore' AS SomeMore
FOR XML PATH('TestPath')
)
set @result.modify('
insert <?xml-stylesheet type="text/xsl" href="stylesheet.xsl"?>
before /*[1]
')
<强> Sqlfiddle Demo
强>
传递给modify()
函数的XQuery表达式告诉SQL Server在XML的根元素之前插入处理指令节点。
更新:
基于以下线程找到另一个替代方案:Merge the two xml fragments into one?。我个人更喜欢这种方式:
SELECT CONVERT(XML, '<?xml-stylesheet type="text/xsl" href="stylesheet.xsl"?>'),
(
SELECT
'Test' AS Test,
'SomeMore' AS SomeMore
FOR XML PATH('TestPath')
)
FOR XML PATH('')
<强> Sqlfiddle Demo
强>
答案 1 :(得分:0)
当它出现时,har07的好答案不适用于XML声明。我能找到的唯一方法是:
DECLARE @ExistingXML XML=
(
SELECT
'Test' AS Test,
'SomeMore' AS SomeMore
FOR XML PATH('TestPath'),TYPE
);
DECLARE @XmlWithDeclaration NVARCHAR(MAX)=
(
SELECT '<?xml version="1.0" encoding="UTF-8"?>'
+
CAST(@ExistingXml AS NVARCHAR(MAX))
);
SELECT @XmlWithDeclaration;
在此步骤之后,您必须留在string
行,任何转换为真实XML 都会出错(当编码为UTF-16
之外的其他编码时)或将省略这个xml声明。