如何按存储在字符串变量中的名称获取 XAML 元素？

Posted 2023-04-12

【中文标题】如何按存储在字符串变量中的名称获取 XAML 元素？

【英文标题】：How to get XAML element by name stored in a string variable?

【发布时间】：2014-03-30 07:49:12

【问题描述】：

例如，我有一个 UIElement：

<TextBlock Name="sometextblock" Text="sample text"/>

在代码中，我有一个具有该名称的字符串变量：

string elementName = "sometextblock";

如何获取这个元素，使用这个变量？我需要访问元素的属性，例如，我需要能够更改 Text 属性。

如何做到这一点？

谢谢！

【参考方案1】：

你可以参考这个方法：

    public static bool FindVisualChildByName<T>(this DependencyObject parent, string name, out T control) where T : DependencyObject
    
        if (parent == null)
            throw new ArgumentNullException(nameof(parent), "Control cấp cha không được null.");

        if (string.IsNullOrWhiteSpace(name))
            throw new ArgumentNullException(nameof(name), "Tên của control cần tìm không được null hoặc empty.");

        var childrenCount = VisualTreeHelper.GetChildrenCount(parent);

        for (var i = 0; i < childrenCount; i++)
        
            var child       = VisualTreeHelper.GetChild(parent, i);
            var controlName = child.GetValue(FrameworkElement.NameProperty) as string;

            if (controlName == name)
            
                control = child as T;
                return true;
            
            if (FindVisualChildByName(child, name, out control)) return true;
        
        control = null;
        return false;
    

【参考方案2】：

假设你的 XAML 上有这个：

<Button Name="MyButton1" .../>
<Button Name="MyButton2" .../>
<Image Name="MyImage1" .../>
<TextBox Name="MyTextBox1" .../>
<TextBox Name="MyTextBox2" .../>
<Button Name="MyButton3" .../>

您必须将控件的名称放在一个字符串数组中，所以：

string[] NameControls =  "MyButton1", "MyButton2", "MyImage1", "MyTextBox1", "MyTextBox2", "MyButton3" ;

然后你可以迭代控件并访问属性：

for (int i = 0; i < NameControls.Length; i++)

    dynamic MyControl = this.FindName(NameControls[i]);
    // do something

例如，在我的情况下，我需要更改已确定控件的不透明度，因此，在 for 块中我已将其放入：

dynamic MyControl = this.FindName(NameControls[i]);
MyControl.Opacity = 0.7;

【参考方案3】：

如果您在 XAML 中按如下方式命名元素：

<TextBlock x:Name="sometextblock" />

您可以通过FindName 方法找到它们：

TextBlock txt = this.FindName("sometextblock") as TextBlock;


string elementName = txt.xyzproperty //do what you want with using txt.xyz property

【参考方案4】：

你可以用这个方法：

var textBlock = FindChild<TextBlock>(Application.Current.RootVisual, "sometextblock");

FindChild 方法是：

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++)
        
            DependencyObject child = VisualTreeHelper.GetChild(parent, i);
            // If the child is not of the request child type child
            var 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;
                

                // Need this in case the element we want is nested
                // in another element of the same type
                foundChild = FindChild<T>(child, childName);
            
            else
            
                // child element found.
                foundChild = (T) child;
                break;
            
        

        return foundChild;
    

【讨论】：

有更简单的方法吗？