如何使用x / y坐标而不是索引访问Grid中的子项？ [英] How to access children in Grid with x/y coordinates instead of index?

问题描述

我可以使用这种语法：

$ b我有一个Grid对象，并希望获得特定的复选框。 $ b

  CheckBox cbChange = grid.Children [4]作为CheckBox; 
  

但是我怎样才能通过x / y坐标来访问这个孩子，例如：

  CheckBox cbChange = grid.GetXYChild（2,3）作为CheckBox; // PSEUDO-CODE 
  

  using System.Collections.Generic; 
使用System.Windows; 
使用System.Windows.Controls; 
使用System.Windows.Documents; 
 
 namespace TestGrid92292 
 {
 public partial class Window1：Window 
 {
 public Window1（）
 {
 InitializeComponent（） ; 
 
列表< string> rowNames =新列表< string> 
 {
营销，
销售，
开发
}; 
 
 Grid grid = new Grid（）; 
 grid.Margin =新厚度（5）; 
 
 ColumnDefinition col1 = new ColumnDefinition（）; 
 col1.Width = new GridLength（100，GridUnitType.Pixel）; 
 grid.ColumnDefinitions.Add（col1）; 
 
 ColumnDefinition col2 = new ColumnDefinition（）; 
 col2.Width = new GridLength（1，GridUnitType.Star）; 
 grid.ColumnDefinitions.Add（col2）; 
 
 ColumnDefinition col3 = new ColumnDefinition（）; 
 col3.Width = new GridLength（1，GridUnitType.Star）; 
 grid.ColumnDefinitions.Add（col3）; 
 
 int rowCount = 0; 
 foreach（rowNames中的var rowName）
 {
 RowDefinition row = new RowDefinition（）; 
 grid.RowDefinitions.Add（row）; 
 
 TextBlock tb = new TextBlock（）; 
 tb.Text = rowName; 
 tb.SetValue（Grid.ColumnProperty，0）; 
 tb.SetValue（Grid.RowProperty，rowCount）; 
 grid.Children.Add（tb）; 
 
 CheckBox cb = new CheckBox（）; 
 cb.SetValue（Grid.ColumnProperty，1）; 
 cb.SetValue（Grid.RowProperty，rowCount）; 
 cb.Horizo​​ntalAlignment = Horizo​​ntalAlignment.Left; 
 grid.Children.Add（cb）; 
 
 CheckBox cb2 = new CheckBox（）; 
 cb2.SetValue（Grid.ColumnProperty，2）; 
 cb2.SetValue（Grid.RowProperty，rowCount）; 
 cb2.Horizo​​ntalAlignment = Horizo​​ntalAlignment.Left; 
 grid.Children.Add（cb2）; 
 
 rowCount ++; 
} 
 
 //检查特定框
 // CheckBox cbChange = grid.GetXYChild（2,3）作为CheckBox; 
 CheckBox cbChange = grid.Children [4]作为CheckBox; 
 cbChange.IsChecked = true; 
 
 MainContent.Children.Add（grid）; 
 
 
 
  

解决方案

我会制作一个扩展方法，它可以获得所需的x值和y值。在这里你通过所有的孩子，并检查Grid.GetRow（child）和Grid.GetColumn（child） - 方法是否返回所需的值。

因此我会返回一个 IEnumerable< FrameworkElement> ，因为你可以有多个，一个或者没有元素在这个位置（不是在你的例子中，而是一般的这种方法）。

类似于：

  public static class GridExtensions {
 public static IEnumerable< FrameworkElement> GetXYChild（这个网格实例，int x，int y）{
 if（null == instance）{
 throw new ArgumentNullException（instance）; 
} 
 List< FrameworkElement> list = new List< FrameworkElement>（）; 
 foreach（instance.Children中的FrameworkElement fe）{
 if（Grid.GetRow（fe）== y&& Grid.GetColumn（fe）== x）{
 list。加入（FE）; 
} 
} 
返回列表; 
} 
} 
  

我没有测试过它，如果发表评论不起作用。如果您只想返回一个实例，则可以根据需要更改代码。

I have a Grid object and want to get a specific checkbox out of it.

I can use this syntax:

CheckBox cbChange = grid.Children[4] as CheckBox;

But how can I access this child via x/y coordinate instead, e.g.:

CheckBox cbChange = grid.GetXYChild(2,3) as CheckBox; //PSEUDO-CODE

using System.Collections.Generic;
using System.Windows;
using System.Windows.Controls;
using System.Windows.Documents;

namespace TestGrid92292
{
    public partial class Window1 : Window
    {
        public Window1()
        {
            InitializeComponent();

            List<string> rowNames = new List<string>
            {
                "Marketing",
                "Sales",
                "Development"
            };

            Grid grid = new Grid();
            grid.Margin = new Thickness(5);

            ColumnDefinition col1 = new ColumnDefinition();
            col1.Width = new GridLength(100, GridUnitType.Pixel);
            grid.ColumnDefinitions.Add(col1);

            ColumnDefinition col2 = new ColumnDefinition();
            col2.Width = new GridLength(1, GridUnitType.Star);
            grid.ColumnDefinitions.Add(col2);

            ColumnDefinition col3 = new ColumnDefinition();
            col3.Width = new GridLength(1, GridUnitType.Star);
            grid.ColumnDefinitions.Add(col3);

            int rowCount = 0;
            foreach (var rowName in rowNames)
            {
                RowDefinition row = new RowDefinition();
                grid.RowDefinitions.Add(row);

                TextBlock tb = new TextBlock();
                tb.Text = rowName;
                tb.SetValue(Grid.ColumnProperty, 0);
                tb.SetValue(Grid.RowProperty, rowCount);
                grid.Children.Add(tb);

                CheckBox cb = new CheckBox();
                cb.SetValue(Grid.ColumnProperty, 1);
                cb.SetValue(Grid.RowProperty, rowCount);
                cb.HorizontalAlignment = HorizontalAlignment.Left;
                grid.Children.Add(cb);

                CheckBox cb2 = new CheckBox();
                cb2.SetValue(Grid.ColumnProperty, 2);
                cb2.SetValue(Grid.RowProperty, rowCount);
                cb2.HorizontalAlignment = HorizontalAlignment.Left;
                grid.Children.Add(cb2);

                rowCount++;
            }

            //check a specific box
            //CheckBox cbChange = grid.GetXYChild(2,3) as CheckBox;
            CheckBox cbChange = grid.Children[4] as CheckBox;
            cbChange.IsChecked = true;

            MainContent.Children.Add(grid);
        }
    }
}

解决方案

I would make an extension method that takes the desired x-and y-values. There in you go through all children and check if the Grid.GetRow(child) and Grid.GetColumn(child)-methods return the desired values.

As a result I would return an IEnumerable<FrameworkElement> because you can have more than one, one or no elements at this position (not in your example, but for such a method in general).

Something like:

public static class GridExtensions {
    public static IEnumerable<FrameworkElement> GetXYChild(this Grid instance, int x, int y) {
        if (null == instance) {
            throw new ArgumentNullException("instance");
        }
        List<FrameworkElement> list = new List<FrameworkElement>();            
        foreach (FrameworkElement fe in instance.Children) {
            if (Grid.GetRow(fe) == y && Grid.GetColumn(fe) == x) {
                list.Add(fe);
            }
        }
        return list;
    }
}

I have not tested it, make a comment if it does not work. If you only want to return one instance, you can change the code acordingly.

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