根据XAM数据网格的名称以编程方式设置所选项目
本文关键字:方式 编程 设置 项目 选项 数据 XAM 数据网 网格 根据 | 更新日期: 2023-09-27 18:09:12
类似这样的内容
private void SearchResult(string nameOfBean)
{
foreach (Record VARIABLE in mbeanDataGrid.Records)
{
if (VARIABLE.ToString().Contains(nameOfBean))
{
((VARIABLE as DataRecord).DataItem as Record).IsSelected = true;
}
}
}
然而,我知道这个语法是错误的,我正在寻找一些建议!基本上是通过代码选择项目(就好像你点击了它一样)。根据其名称
您可以使用以下代码选择记录(如果您想要选择多个记录)
private void ShowSearchResult(string searchStr)
{
var recordsToSelect = new List<Record>();
foreach (Record rec in xamGrid.Records) {
var yourData = rec is DataRecord ? ((DataRecord)rec).DataItem as YourDataClass : null;
if (yourData != null && yourData.MatchWithSearchStr(searchStr)) {
recordsToSelect.Add(rec);
}
}
xamGrid.SelectedItems.Records.Clear();
// you need linq -> .ToArray()
xamGrid.SelectedItems.Records.AddRange(recordsToSelect.ToArray(), false, true);
}
或者如果你只想激活和选择一个记录,那么执行这个
private void ShowSearchResult(string searchStr)
{
foreach (Record rec in xamGrid.Records) {
var yourData = rec is DataRecord ? ((DataRecord)rec).DataItem as YourDataClass : null;
if (yourData != null && yourData.MatchWithSearchStr(searchStr)) {
xamGrid.ActiveRecord = rec;
// don't know if you really need this
xamGrid.ActiveRecord.IsSelected = true;
break;
}
}
}
希望能有所帮助