System.Windows.Forms.TextBox不会更改引用文本

时间:2018-09-06 19:36:52

标签: c# winforms dialog inputbox

我有以下代码作为弹出对话框,该对话框从inputBox接收输入。我传入字符串作为参考,希望参考字符串在对话框关闭时会改变,因此我可以获取用户输入。但是传入的字符串在对话框关闭时没有改变。我做错了什么?

public static DialogResult ShowInputDialog(ref string input1, ref string input2)
{
    var size = new System.Drawing.Size(520, 180);
    var inputBox = new Form { ClientSize = size };

    var panel = new TableLayoutPanel
    {
        Size = new System.Drawing.Size(460, 180),
        Location = new System.Drawing.Point(25, 15),
        ColumnCount = 2,
        RowCount = 3
    };

    // Add ColumnStyles/RowStyles here

    panel.Controls.Add(new Label { Text = "Input 1", TextAlign = ContentAlignment.BottomRight }, 0, 0);
    panel.Controls.Add(new Label { Text = "Input2", TextAlign = ContentAlignment.BottomRight }, 0, 1);
    panel.Controls.Add(new TextBox { Text = input1, Width = 280 }, 1, 0);
    panel.Controls.Add(new TextBox { Text = input2, Width = 280 }, 1, 1);
    var okButton = new Button{ DialogResult = DialogResult.OK};
    var cancelButton = new Button {DialogResult = DialogResult.Cancel};

    var buttons = new FlowLayoutPanel();
    buttons.Controls.Add(okButton);
    buttons.Controls.Add(cancelButton);
    panel.Controls.Add(buttons, 1, 3);
    inputBox.Controls.Add(panel);

    inputBox.AcceptButton = okButton;
    inputBox.CancelButton = cancelButton;

    var result = inputBox.ShowDialog();
    return result;
}

上面代码的用法是:

string input1 = string.Empty; 
string input2 = string.Empty;
ShowInputDialog(ref input, ref input2);

2 个答案:

答案 0 :(得分:1)

用户单击确定按钮后,必须将textbox.text值分配回input1和input2

答案 1 :(得分:0)

我对TableLayoutPanel不太熟悉,但是也许您可以做一些简单的事情:

 if (inputBox.ShowDialog() == DialogResult.OK)
 {
    input1 = (panel.GetControlFromPosition(1, 0) as TextBox).Text;
    input2 = (panel.GetControlFromPosition(1, 1) as TextBox).Text;
    return DialogResult.OK;
 }

 return DialogResult.Cancel;

当前您的问题是,在关闭对话框之后,您实际上没有在任何地方设置该值。

但是,我同意这一评论。一种MVVM模式可能会使这些类型的属性及其相应值的维护(和创建)更加容易。