我需要在WPF控件层次结构中搜索与给定名称或类型匹配的控件。我怎么能这样做？
答案 0 :(得分：292)
我结合了John Myczek使用的模板格式和上面的Tri Q算法来创建一个可以在任何父级上使用的findChild算法。请记住，向下递归搜索树可能是一个漫长的过程。我只是在WPF应用程序上对此进行了检查，请对您可能发现的任何错误发表评论，我将更正我的代码。
WPF Snoop是查看可视化树的有用工具 - 我强烈建议在测试或使用此算法检查您的工作时使用它。
Tri Q算法中存在一个小错误。找到子项后，如果childrenCount为＆gt; 1，我们再次迭代，我们可以覆盖正确找到的孩子。因此，我在代码中添加了if (foundChild != null) break;来处理这种情况。
/// <summary>
/// Finds a Child of a given item in the visual tree.
/// </summary>
/// <param name="parent">A direct parent of the queried item.</param>
/// <typeparam name="T">The type of the queried item.</typeparam>
/// <param name="childName">x:Name or Name of child. </param>
/// <returns>The first parent item that matches the submitted type parameter.
/// If not matching item can be found,
/// a null parent is being returned.</returns>
public static T FindChild<T>(DependencyObject parent, string childName)
where T : DependencyObject
{
// Confirm parent and childName are valid.
if (parent == null) return null;
T foundChild = null;
int childrenCount = VisualTreeHelper.GetChildrenCount(parent);
for (int i = 0; i < childrenCount; i++)
{
var child = VisualTreeHelper.GetChild(parent, i);
// If the child is not of the request child type child
T childType = child as T;
if (childType == null)
{
// recursively drill down the tree
foundChild = FindChild<T>(child, childName);
// If the child is found, break so we do not overwrite the found child.
if (foundChild != null) break;
}
else if (!string.IsNullOrEmpty(childName))
{
var frameworkElement = child as FrameworkElement;
// If the child's name is set for search
if (frameworkElement != null && frameworkElement.Name == childName)
{
// if the child's name is of the request name
foundChild = (T)child;
break;
}
}
else
{
// child element found.
foundChild = (T)child;
break;
}
}
return foundChild;
}
这样称呼：
TextBox foundTextBox =
UIHelper.FindChild<TextBox>(Application.Current.MainWindow, "myTextBoxName");
注意Application.Current.MainWindow可以是任何父窗口。
答案 1 :(得分：117)
您还可以使用FrameworkElement.FindName(string)按名称查找元素。
假设：
<UserControl ...>
<TextBlock x:Name="myTextBlock" />
</UserControl>
在代码隐藏文件中，您可以编写：
var myTextBlock = (TextBlock)this.FindName("myTextBlock");
当然，因为它是使用x：Name定义的，所以你可以引用生成的字段，但也许你想要动态查找它而不是静态查找。
此方法也可用于模板，其中命名项目多次出现（每次使用模板一次）。
答案 2 :(得分：64)
您可以使用VisualTreeHelper查找控件。下面是一个使用VisualTreeHelper查找指定类型的父控件的方法。您也可以使用VisualTreeHelper以其他方式查找控件。
public static class UIHelper
{
/// <summary>
/// Finds a parent of a given item on the visual tree.
/// </summary>
/// <typeparam name="T">The type of the queried item.</typeparam>
/// <param name="child">A direct or indirect child of the queried item.</param>
/// <returns>The first parent item that matches the submitted type parameter.
/// If not matching item can be found, a null reference is being returned.</returns>
public static T FindVisualParent<T>(DependencyObject child)
where T : DependencyObject
{
// get parent item
DependencyObject parentObject = VisualTreeHelper.GetParent(child);
// we’ve reached the end of the tree
if (parentObject == null) return null;
// check if the parent matches the type we’re looking for
T parent = parentObject as T;
if (parent != null)
{
return parent;
}
else
{
// use recursion to proceed with next level
return FindVisualParent<T>(parentObject);
}
}
}
这样称呼：
Window owner = UIHelper.FindVisualParent<Window>(myControl);
答案 3 :(得分：20)
我可能只是在重复其他人，但我确实有一段很好的代码，它使用FindChild（）方法扩展DependencyObject类，它将按类型和名称为您提供子类。只需包括并使用。
public static class UIChildFinder
{
public static DependencyObject FindChild(this DependencyObject reference, string childName, Type childType)
{
DependencyObject foundChild = null;
if (reference != null)
{
int childrenCount = VisualTreeHelper.GetChildrenCount(reference);
for (int i = 0; i < childrenCount; i++)
{
var child = VisualTreeHelper.GetChild(reference, i);
// If the child is not of the request child type child
if (child.GetType() != childType)
{
// recursively drill down the tree
foundChild = FindChild(child, childName, childType);
}
else if (!string.IsNullOrEmpty(childName))
{
var frameworkElement = child as FrameworkElement;
// If the child's name is set for search
if (frameworkElement != null && frameworkElement.Name == childName)
{
// if the child's name is of the request name
foundChild = child;
break;
}
}
else
{
// child element found.
foundChild = child;
break;
}
}
}
return foundChild;
}
}
希望你觉得它很有用。
答案 4 :(得分：18)
我对代码的扩展。
来源： https://code.google.com/p/gishu-util/source/browse/#git%2FWPF%2FUtilities
解释性博文： http://madcoderspeak.blogspot.com/2010/04/wpf-find-child-control-of-specific-type.html
答案 5 :(得分：18)
如果您想查找特定类型的所有控件，您可能也对此代码段感兴趣
public static IEnumerable<T> FindVisualChildren<T>(DependencyObject parent)
where T : DependencyObject
{
int childrenCount = VisualTreeHelper.GetChildrenCount(parent);
for (int i = 0; i < childrenCount; i++)
{
var child = VisualTreeHelper.GetChild(parent, i);
var childType = child as T;
if (childType != null)
{
yield return (T)child;
}
foreach (var other in FindVisualChildren<T>(child))
{
yield return other;
}
}
}
答案 6 :(得分：16)
我编辑了CrimsonX的代码，因为它没有使用超类类型：
public static T FindChild<T>(DependencyObject depObj, string childName)
where T : DependencyObject
{
// Confirm obj is valid.
if (depObj == null) return null;
// success case
if (depObj is T && ((FrameworkElement)depObj).Name == childName)
return depObj as T;
for (int i = 0; i < VisualTreeHelper.GetChildrenCount(depObj); i++)
{
DependencyObject child = VisualTreeHelper.GetChild(depObj, i);
//DFS
T obj = FindChild<T>(child, childName);
if (obj != null)
return obj;
}
return null;
}
答案 7 :(得分：15)
这将忽略一些元素 - 您应该像这样扩展它以支持更广泛的控件。有关简短讨论，请查看here
/// <summary>
/// Helper methods for UI-related tasks.
/// </summary>
public static class UIHelper
{
/// <summary>
/// Finds a parent of a given item on the visual tree.
/// </summary>
/// <typeparam name="T">The type of the queried item.</typeparam>
/// <param name="child">A direct or indirect child of the
/// queried item.</param>
/// <returns>The first parent item that matches the submitted
/// type parameter. If not matching item can be found, a null
/// reference is being returned.</returns>
public static T TryFindParent<T>(DependencyObject child)
where T : DependencyObject
{
//get parent item
DependencyObject parentObject = GetParentObject(child);
//we've reached the end of the tree
if (parentObject == null) return null;
//check if the parent matches the type we're looking for
T parent = parentObject as T;
if (parent != null)
{
return parent;
}
else
{
//use recursion to proceed with next level
return TryFindParent<T>(parentObject);
}
}
/// <summary>
/// This method is an alternative to WPF's
/// <see cref="VisualTreeHelper.GetParent"/> method, which also
/// supports content elements. Do note, that for content element,
/// this method falls back to the logical tree of the element!
/// </summary>
/// <param name="child">The item to be processed.</param>
/// <returns>The submitted item's parent, if available. Otherwise
/// null.</returns>
public static DependencyObject GetParentObject(DependencyObject child)
{
if (child == null) return null;
ContentElement contentElement = child as ContentElement;
if (contentElement != null)
{
DependencyObject parent = ContentOperations.GetParent(contentElement);
if (parent != null) return parent;
FrameworkContentElement fce = contentElement as FrameworkContentElement;
return fce != null ? fce.Parent : null;
}
//if it's not a ContentElement, rely on VisualTreeHelper
return VisualTreeHelper.GetParent(child);
}
}
答案 8 :(得分：12)
虽然我喜欢一般的递归，但是当用C＃编程时，它不如迭代效率高，所以下面的解决方案可能比John Myczek建议的更简洁？这将从给定控件中搜索层次结构，以查找特定类型的祖先控件。
public static T FindVisualAncestorOfType<T>(this DependencyObject Elt)
where T : DependencyObject
{
for (DependencyObject parent = VisualTreeHelper.GetParent(Elt);
parent != null; parent = VisualTreeHelper.GetParent(parent))
{
T result = parent as T;
if (result != null)
return result;
}
return null;
}
像这样调用它来查找包含名为Window的控件的ExampleTextBox：
Window window = ExampleTextBox.FindVisualAncestorOfType<Window>();
答案 9 :(得分：9)
exciton80 ...我的代码没有通过usercontrols递归出现问题。它击中了Grid根并抛出错误。我相信这可以解决它：
public static object[] FindControls(this FrameworkElement f, Type childType, int maxDepth)
{
return RecursiveFindControls(f, childType, 1, maxDepth);
}
private static object[] RecursiveFindControls(object o, Type childType, int depth, int maxDepth = 0)
{
List<object> list = new List<object>();
var attrs = o.GetType().GetCustomAttributes(typeof(ContentPropertyAttribute), true);
if (attrs != null && attrs.Length > 0)
{
string childrenProperty = (attrs[0] as ContentPropertyAttribute).Name;
if (String.Equals(childrenProperty, "Content") || String.Equals(childrenProperty, "Children"))
{
var collection = o.GetType().GetProperty(childrenProperty).GetValue(o, null);
if (collection is System.Windows.Controls.UIElementCollection) // snelson 6/6/11
{
foreach (var c in (IEnumerable)collection)
{
if (c.GetType().FullName == childType.FullName)
list.Add(c);
if (maxDepth == 0 || depth < maxDepth)
list.AddRange(RecursiveFindControls(
c, childType, depth + 1, maxDepth));
}
}
else if (collection != null && collection.GetType().BaseType.Name == "Panel") // snelson 6/6/11; added because was skipping control (e.g., System.Windows.Controls.Grid)
{
if (maxDepth == 0 || depth < maxDepth)
list.AddRange(RecursiveFindControls(
collection, childType, depth + 1, maxDepth));
}
}
}
return list.ToArray();
}
答案 10 :(得分：9)
这是我的代码，用于按类型查找控件，同时控制我们进入层次结构的深度 （maxDepth == 0表示无限深）。
public static class FrameworkElementExtension
{
public static object[] FindControls(
this FrameworkElement f, Type childType, int maxDepth)
{
return RecursiveFindControls(f, childType, 1, maxDepth);
}
private static object[] RecursiveFindControls(
object o, Type childType, int depth, int maxDepth = 0)
{
List<object> list = new List<object>();
var attrs = o.GetType()
.GetCustomAttributes(typeof(ContentPropertyAttribute), true);
if (attrs != null && attrs.Length > 0)
{
string childrenProperty = (attrs[0] as ContentPropertyAttribute).Name;
foreach (var c in (IEnumerable)o.GetType()
.GetProperty(childrenProperty).GetValue(o, null))
{
if (c.GetType().FullName == childType.FullName)
list.Add(c);
if (maxDepth == 0 || depth < maxDepth)
list.AddRange(RecursiveFindControls(
c, childType, depth + 1, maxDepth));
}
}
return list.ToArray();
}
}
答案 11 :(得分：8)
我有一个像这样的序列函数（这是完全一般的）：
public static IEnumerable<T> SelectAllRecursively<T>(this IEnumerable<T> items, Func<T, IEnumerable<T>> func)
{
return (items ?? Enumerable.Empty<T>()).SelectMany(o => new[] { o }.Concat(SelectAllRecursively(func(o), func)));
}
立即生孩子：
public static IEnumerable<DependencyObject> FindChildren(this DependencyObject obj)
{
return Enumerable.Range(0, VisualTreeHelper.GetChildrenCount(obj))
.Select(i => VisualTreeHelper.GetChild(obj, i));
}
在分层树上寻找所有孩子：
public static IEnumerable<DependencyObject> FindAllChildren(this DependencyObject obj)
{
return obj.FindChildren().SelectAllRecursively(o => o.FindChildren());
}
你可以在Window上调用它来获取所有控件。
收集完成后，可以使用LINQ（即OfType，Where）。
答案 12 :(得分：6)
由于问题很普遍，可能会吸引人们寻找非常微不足道的案例的答案：如果你只想要一个孩子而不是一个后代，你可以使用Linq：
private void ItemsControlItem_Loaded(object sender, RoutedEventArgs e)
{
if (SomeCondition())
{
var children = (sender as Panel).Children;
var child = (from Control child in children
where child.Name == "NameTextBox"
select child).First();
child.Focus();
}
}
或当然是循环迭代儿童的明显for循环。
答案 13 :(得分：3)
这些选项已经讨论过在C＃中遍历Visual Tree。 它也可以使用RelativeSource标记扩展来遍历xaml中的可视化树。 msdn
按类型查找
Binding="{Binding RelativeSource={RelativeSource Mode=FindAncestor, AncestorType={x:Type <TypeToFind>}}}"
答案 14 :(得分：2)
这是一个使用灵活谓词的解决方案：
public static DependencyObject FindChild(DependencyObject parent, Func<DependencyObject, bool> predicate)
{
if (parent == null) return null;
int childrenCount = VisualTreeHelper.GetChildrenCount(parent);
for (int i = 0; i < childrenCount; i++)
{
var child = VisualTreeHelper.GetChild(parent, i);
if (predicate(child))
{
return child;
}
else
{
var foundChild = FindChild(child, predicate);
if (foundChild != null)
return foundChild;
}
}
return null;
}
您可以这样称呼它：
var child = FindChild(parent, child =>
{
var textBlock = child as TextBlock;
if (textBlock != null && textBlock.Name == "MyTextBlock")
return true;
else
return false;
}) as TextBlock;
答案 15 :(得分：1)
此代码只修复了@CrimsonX回答的错误：
FrameworkElement如果类型匹配但名称不合适，则需要以递归方式继续调用该方法（当您将T作为null传递时会发生这种情况）。否则它会返回mysql> SHOW COLLATION LIKE '%latin1%';
+-------------------+---------+----+---------+----------+---------+
| Collation | Charset | Id | Default | Compiled | Sortlen |
+-------------------+---------+----+---------+----------+---------+
| latin1_german1_ci | latin1 | 5 | | Yes | 1 |
| latin1_swedish_ci | latin1 | 8 | Yes | Yes | 1 |
| latin1_danish_ci | latin1 | 15 | | Yes | 1 |
| latin1_german2_ci | latin1 | 31 | | Yes | 2 |
| latin1_bin | latin1 | 47 | | Yes | 1 |
| latin1_general_ci | latin1 | 48 | | Yes | 1 |
| latin1_general_cs | latin1 | 49 | | Yes | 1 |
| latin1_spanish_ci | latin1 | 94 | | Yes | 1 |
+-------------------+---------+----+---------+----------+---------+
而这是错误的。
答案 16 :(得分：1)
我无法使用 @CrimsonX 或 @Drew Noakes 方法找到我的控件，因为我的 ControlTemplate 位于单独的 XAML 文件中。 我使用以下方法找到了控件：
private Image backImage;
private void Diagram_Loaded(object sender, System.Windows.RoutedEventArgs e)
{
// here is the example of the ControlTemplate declaration
//<ControlTemplate x:Key="DiagramControlTemplate1" TargetType="{x:Type Diagram}">
var ctrlTemplate = (ControlTemplate)FindResource("DiagramControlTemplate1");
// diagram is the x:Name of TemplatedControl and, "backImage" is the name of control that I want to find.
var imageControl = ctrlTemplate.FindName("backImage", diagram);
if (imageControl != null)
{
this.backImage = (Image)imageControl;
}
}
这对我有用。
答案 17 :(得分：0)
要从代码中查找给定类型的祖先，您可以使用：
[CanBeNull]
public static T FindAncestor<T>(DependencyObject d) where T : DependencyObject
{
while (true)
{
d = VisualTreeHelper.GetParent(d);
if (d == null)
return null;
var t = d as T;
if (t != null)
return t;
}
}
此实现使用迭代而不是递归，这可能会稍快一些。
如果您使用的是C＃7，可以稍微缩短一下：
[CanBeNull]
public static T FindAncestor<T>(DependencyObject d) where T : DependencyObject
{
while (true)
{
d = VisualTreeHelper.GetParent(d);
if (d == null)
return null;
if (d is T t)
return t;
}
}
答案 18 :(得分：-5)
试试这个
<TextBlock x:Name="txtblock" FontSize="24" >Hai Welcom to this page
</TextBlock>
背后的代码
var txtblock = sender as Textblock;
txtblock.Foreground = "Red"