SyntaxNode.DescendantNodesAndTokens 方法
定义
重要
一些信息与预发行产品相关,相应产品在发行之前可能会进行重大修改。 对于此处提供的信息,Microsoft 不作任何明示或暗示的担保。
重载
DescendantNodesAndTokens(Func<SyntaxNode,Boolean>, Boolean) |
按前缀文档顺序获取子代节点和令牌的列表。 |
DescendantNodesAndTokens(TextSpan, Func<SyntaxNode,Boolean>, Boolean) |
按前缀文档顺序获取子代节点和令牌的列表。 |
DescendantNodesAndTokens(Func<SyntaxNode,Boolean>, Boolean)
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
- Source:
- SyntaxNode.cs
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- SyntaxNode.cs
- Source:
- SyntaxNode.cs
按前缀文档顺序获取子代节点和令牌的列表。
public System.Collections.Generic.IEnumerable<Microsoft.CodeAnalysis.SyntaxNodeOrToken> DescendantNodesAndTokens (Func<Microsoft.CodeAnalysis.SyntaxNode,bool> descendIntoChildren = default, bool descendIntoTrivia = false);
public System.Collections.Generic.IEnumerable<Microsoft.CodeAnalysis.SyntaxNodeOrToken> DescendantNodesAndTokens (Func<Microsoft.CodeAnalysis.SyntaxNode,bool>? descendIntoChildren = default, bool descendIntoTrivia = false);
member this.DescendantNodesAndTokens : Func<Microsoft.CodeAnalysis.SyntaxNode, bool> * bool -> seq<Microsoft.CodeAnalysis.SyntaxNodeOrToken>
Public Function DescendantNodesAndTokens (Optional descendIntoChildren As Func(Of SyntaxNode, Boolean) = Nothing, Optional descendIntoTrivia As Boolean = false) As IEnumerable(Of SyntaxNodeOrToken)
参数
- descendIntoChildren
- Func<SyntaxNode,Boolean>
一个可选函数,用于确定搜索是否进入参数节点的子级。
- descendIntoTrivia
- Boolean
确定属于结构化琐事的节点是否包含在列表中。
返回
适用于
DescendantNodesAndTokens(TextSpan, Func<SyntaxNode,Boolean>, Boolean)
- Source:
- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
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- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
- Source:
- SyntaxNode.cs
按前缀文档顺序获取子代节点和令牌的列表。
public System.Collections.Generic.IEnumerable<Microsoft.CodeAnalysis.SyntaxNodeOrToken> DescendantNodesAndTokens (Microsoft.CodeAnalysis.Text.TextSpan span, Func<Microsoft.CodeAnalysis.SyntaxNode,bool> descendIntoChildren = default, bool descendIntoTrivia = false);
public System.Collections.Generic.IEnumerable<Microsoft.CodeAnalysis.SyntaxNodeOrToken> DescendantNodesAndTokens (Microsoft.CodeAnalysis.Text.TextSpan span, Func<Microsoft.CodeAnalysis.SyntaxNode,bool>? descendIntoChildren = default, bool descendIntoTrivia = false);
member this.DescendantNodesAndTokens : Microsoft.CodeAnalysis.Text.TextSpan * Func<Microsoft.CodeAnalysis.SyntaxNode, bool> * bool -> seq<Microsoft.CodeAnalysis.SyntaxNodeOrToken>
Public Function DescendantNodesAndTokens (span As TextSpan, Optional descendIntoChildren As Func(Of SyntaxNode, Boolean) = Nothing, Optional descendIntoTrivia As Boolean = false) As IEnumerable(Of SyntaxNodeOrToken)
参数
- span
- TextSpan
节点的完整跨度必须相交。
- descendIntoChildren
- Func<SyntaxNode,Boolean>
一个可选函数,用于确定搜索是否进入参数节点的子级。
- descendIntoTrivia
- Boolean
确定属于结构化琐事的节点是否包含在列表中。