Question

我写一个程序，有必要在其中切换当前状态，有必要在计划事件时计划它，它被认为是一个对象，该对象具有自己的字段，例如事件的开始时间和结束时间，我希望在生成工作表装箱时输出此对象。告诉我该怎么做？

List<ChangeStatus> events = new List<ChangeStatus>();

private void toPlanButton_Click(object sender, EventArgs e)
{
    string comboBoxTypeNumber = comboBoxType.SelectedItem.ToString();

    DateTime Time = new DateTime();
    Time = dateTimePicker1.Value;
    DateTime longTime = new DateTime();
    longTime = dateTimePicker2.Value;

    ChangeStatus statusEvent = new ChangeStatus();
    statusEvent.StartEvent = Time;
    statusEvent.LongEvent = longTime;
    statusEvent.TypeEvent = comboBoxTypeNumber;
    events.Add(statusEvent);

    TimeComparer tc = new TimeComparer();
    events.Sort(tc);
}

如何在列表框中显示对象？必须显示对象列表，因为将来我要编辑对象

listBoxEvent.Items.Add("type: " + statusEvent.TypeEvent + ";" + " start: " + statusEvent.StartEvent + ";" + " long: " + statusEvent.LongEvent + " min;"); - work

Answer 1

您可以使用System.Linq Linq来获取字符串文本，并可以在AddRange()集合上调用Items方法，例如

List<string> listData = events.Select(x => "type: " + x.TypeEvent + ";" + " start: " + x.StartEvent + ";" + " long: " + x.LongEvent + " min;").ToList();

listBoxEvent.DataSource = listData;

c＃将列表项传输到列表框

1 个答案: