在listpicker wp8中显示所选值-Java 学习之路

我想在listpicker中显示选定的值 .

Location": [
                {
                    "id": "208",
                    "Name": "Canberra"
                },
                {
                    "id": "209",
                    "Name": "Regional ACT"
                },
                {
                    "id": "67",
                    "Name": "NSW"
                },
                {
                    "id": "134",
                    "Name": "CBD, Inner West & Eastern Suburbs"
                }
            ],

我将此数据转换为list.Now我想将所选位置显示为listpicker

foreach (var Seletedloc in _lst)
{
lstLocations.SelectedItem = Seletedloc ;
}

但我收到此错误.SelectedItem必须始终设置为有效值

2 回答

示例XAML

<Grid Loaded="ContentPanel_OnLoaded" x:Name="ContentPanel" Grid.Row="1"                Margin="12,0,12,0">
        <toolkit:ListPicker Name="MyListPicker"></toolkit:ListPicker>
    </Grid>

示例代码：

private void ContentPanel_OnLoaded(object sender, RoutedEventArgs e)
    {
        var datasource = new[] {"one", "two", "three"};
        MyListPicker.ItemsSource = datasource;
        MyListPicker.SelectedItem = datasource[1];
    }

回复于 2024-04-30T18:14:45+08:00

0
将形成的列表分配给列表选择器的ItemsSource属性并设置所选项 .
```
MyListPicker.ItemsSource = LocationList;
var item = LocationList[1];
MyListPicker.SelectedItem = item;
```
要么
```
MyListPicker.SelectedIndex = 1;
```
如果SelectedItem为null或者ListPicker的ItemsSource中不存在分配给SelectedItem属性的值，则ListPicker将显示此错误 .
回复于 2024-04-30T18:14:45+08:00

在listpicker wp8中显示所选值

2 回答

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