Question

好的，我有一个提供的自定义XML字符串，因此无法修改结构，它是一个键值对，如此处所示

<?xml version='1.0' encoding='utf-8'?>
<ContentStack ID='13' Type='11'>
    <ContentElement1 ID='11' Type='IMAGE'>
       <Key AttributeName='Name' AttributeValue='SOMENAME'></Key>
       <Key AttributeName='SRC' AttributeValue='SOMEFILE.jpg'></Key>
       <Key AttributeName='ALT' AttributeValue='SOMEALT'></Key>
       <Key AttributeName='Class' AttributeValue='SOMECLASS'></Key>
   </ContentElement1>
</ContentStack>

我希望能够抓住attribute value并将它们分配给变量，所以我最终会得到类似的结果。

Dim thsValueOne AS String = "SOMENAME"
Dim thsValueTwo AS String = "SOMEFILE.jpg"
Dim thsValueThree AS String = "SOMEALT"
Dim thsValueFour AS String = "SOMECLASS"

我尝试过这样做但没有运气

Dim xDoc As XDocument = XDocument.Parse(" MY XML STING IN HERE ")
thsValueOne = xDoc.Descendants("Key.AttributeValue").Skip(0).Take(1).ToString

我想我不在附近。

Answer 1

好的，能够根据问题得出这个基于XML的答案很简单

Dim xDoc As XDocument = XDocument.Parse(thsXmlContent)
thsValueOne = xDoc.Root.Descendants.<Key>(0).@AttributeValue
thsValueTwo = xDoc.Root.Descendants.<Key>(1).@AttributeValue
thsValueThree = xDoc.Root.Descendants.<Key>(2).@AttributeValue
thsValueFour = xDoc.Root.Descendants.<Key>(3).@AttributeValue

Answer 2

您可以尝试以下任何一种解决方法，以便了解这一概念。

当您使用VB.NET时，您可以使用XML文字：

Private Sub test1()
    Dim xFrag = <?xml version='1.0' encoding='utf-8'?>
                <ContentStack ID='13' Type='11'>
                    <ContentElement1 ID='11' Type='IMAGE'>
                        <Key AttributeName='Name' AttributeValue='SOMENAME'></Key>
                        <Key AttributeName='SRC' AttributeValue='SOMEFILE.jpg'></Key>
                        <Key AttributeName='ALT' AttributeValue='SOMEALT'></Key>
                        <Key AttributeName='Class' AttributeValue='SOMECLASS'></Key>
                    </ContentElement1>
                </ContentStack>
    Dim query = xFrag...<Key>, pos& = 0
    For Each ele As XElement In query
        pos += 1
        MsgBox(String.Format("Test1:Key({0}):AttributeName='{1}' AttributeValue='{2}'", pos.ToString, ele.@AttributeName, ele.@AttributeValue))
    Next ele
End Sub

或以其他方式使用Parse方法。

Private Sub test2()
    Dim xStr = "<?xml version='1.0' encoding='utf-8'?><ContentStack ID='13' Type='11'><ContentElement1 ID='11' Type='IMAGE'><Key AttributeName='Name' AttributeValue='SOMENAME'></Key><Key AttributeName='SRC' AttributeValue='SOMEFILE.jpg'></Key><Key AttributeName='ALT' AttributeValue='SOMEALT'></Key><Key AttributeName='Class' AttributeValue='SOMECLASS'></Key></ContentElement1></ContentStack>"
    Dim xFrag = XDocument.Parse(xStr)
    Dim query = xFrag...<Key>, pos& = 0
    For Each ele As XElement In query
        pos += 1
        MsgBox(String.Format("Test2:Key({0}):AttributeName='{1}' AttributeValue='{2}'", pos.ToString, ele.@AttributeName, ele.@AttributeValue))
    Next ele
End Sub

希望这会有所帮助......

在VB中使用asp.NET读取自定义XML String

2 个答案: