使用用户控件弹出窗口进行过滤器绑定 [英] Using User Control popup for filter bindingSource

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问题描述

我有一个用户控件,其中有一个checkedListBox和一个按钮.我在dataGridView1_CellDoubleClick事件(我将在下面发布代码)上的弹出按钮单击我想使用bindingSource.Filter"时显示此弹出窗口.

我从此用户控件button_click事件尝试了此操作,但是它什么也没做,于是我以其中存在datagridview却仍然什么都没有的形式创建了一个方法.因此,我需要有关如何获取列名的帮助,而不是需要可以过滤bindingSource的make方法的帮助.

请帮我.我正在使用C#3.5 WinForms

这是用户控制按钮的代码:

Hi, I have a user control with a checkedListBox and one button. I show this fill this popup on dataGridView1_CellDoubleClick event (I''ll post code below) on popup button click I want to use bindingSource.Filter.

I tried it from this user control button_click event but it was doing nothing then I made a method in the form where there is the datagridview and still nothing. So I need help on how to get column name and than make method which can filter bindingSource.

Please help me. I am using C# 3.5 WinForms

This is the user control button code:

private void button1_Click(object sender, EventArgs e)
    {
        OperatorsForm of = new OperatorsForm();
        var checkedvalues = checkedListBox1.CheckedItems.Cast<string>().Select(s => "'" + s + "%'").ToArray();
        string  d = "LoadName LIKE " + string.Join(" or ", checkedvalues);
        of.filter(d);
    }


这是弹出代码:


This is popup code :

private void dataGridView1_CellDoubleClick_1(object sender, DataGridViewCellEventArgs e)
    {
        UserControl1 ucontrol = new UserControl1();
        var button = sender as DataGridView;

        HashSet<string> strings = new HashSet<string>();
        for (int i = 0; i < dataGridView1.Rows.Count - 1; i++)
        {
            strings.Add(dataGridView1.Rows[i].Cells[e.ColumnIndex].Value.ToString());
        }

        checkedListBox1.Items.AddRange(strings.ToArray());
        ucontrol.FillList(strings.ToList());

        var listViewItems = strings.Select(x => new ListViewItem(x, 0)).ToArray();

        ListView listView = new ListView();
        listView.View = View.SmallIcon;
        listView.MultiSelect = false;

        listView.Items.AddRange(listViewItems);

        int itemToShow = 18;
        var lastItemToShow = listViewItems.Take(itemToShow).Last();
        int height = lastItemToShow.Bounds.Bottom + listView.Margin.Top;
        listView.Height = height;
        var popup = new ToolStripDropDown();
        popup.AutoSize = false;
        popup.Margin = Padding.Empty;
        popup.Padding = Padding.Empty;
        ToolStripControlHost host = new ToolStripControlHost(ucontrol);
        host.Margin = Padding.Empty;
        host.Padding = Padding.Empty;
        host.AutoSize = false;
        host.Size = ucontrol.Size;
        popup.Size = ucontrol.Size;
        popup.Items.Add(host);

        popup.Show(this, dataGridView1.Top, dataGridView1.Top);
    }

推荐答案

我认为您的问题出在button1_Click事件处理程序中

您的代码:
Where I think that your problem lies is in the button1_Click event handler

Your Code:
private void button1_Click(object sender, EventArgs e)
    {
        OperatorsForm of = new OperatorsForm(); //<==== #1 =========
        var checkedvalues = checkedListBox1.CheckedItems.Cast<string>().Select(s => "'" + s + "%'").ToArray();
        string  d = "LoadName LIKE " + string.Join(" or ", checkedvalues);
        of.filter(d);  //<=========== #2 =================
    }
</string>



我已对问题行进行编号.

1.这将创建Form的新实例,它 not 不会获得对现有Form的引用.对其进行如下修改:



I have numbered the problem lines.

1. This creates a new instance of the Form, it does not get a reference to the existing Form. Modify it as follows:

OperatorsForm of = this.FindForm() as OperatorsForm


<罢工> 2.删除第1行之后.of不再存在.现在要做什么取决于您是否具有filter(string filterString)方法(a.)(b.).
一个.使用this.filter(d);
b.使用this.myBindingSource.Filter = d;

现在,您应该可以像以前一样使用of.filter(d);.

希望我了解您的问题,对您有所帮助.

祝你好运. :)


2. After deleting line 1. of no longer exists. What you do now depends on whether you have a filter(string filterString) method(a.) or not (b.).
a. use this.filter(d);
b. use this.myBindingSource.Filter = d;

You should now be able to use of.filter(d); as before.

I hope that I have understood your problem and that this helps.

Good luck. :)


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