无法在 datagridview 中打开浏览
如图所示，我有一个 DataGridView，当我单击“从文件浏览”时，必须打开 OpenFileDialog。我相信如果我在 Button_Click 方法下编写代码，但我不知道在哪种方法下编写代码，则没有办法成为可能。请帮忙。
到目前为止我已经做到了：（希望有帮助）
string[] col2 = new string[dataGridView1.Rows.Count];
for (int i = 0; i < dataGridView1.Rows.Count; i++)
if (col2[i] == "Browse From File...")
{
DialogResult result2 = openFileDialog2.ShowDialog();
if (result2 == DialogResult.OK)
{
filename = openFileDialog1.FileName;
}
}
I have a DataGridView as seen in the picture and a OpenFileDialog has to open when I click on "Browse from file". I believe there is no way to become possible if i write my code under the button_Click methods but i dont know under which method to write my code. Please help.
As far as now I have made this: (hope it helps)
string[] col2 = new string[dataGridView1.Rows.Count];
for (int i = 0; i < dataGridView1.Rows.Count; i++)
if (col2[i] == "Browse From File...")
{
DialogResult result2 = openFileDialog2.ShowDialog();
if (result2 == DialogResult.OK)
{
filename = openFileDialog1.FileName;
}
}
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将代码放入
DataGridView.CellValueChanged事件中。使用SelectedCells属性查找当前单元格，然后检查其值是否为“从文件浏览...”。如果是这样，请关闭OpenFileDialog.ShowDialog()调用。有关 CellValueChanged，请参阅此文档： http://msdn .microsoft.com/en-us/library/system.windows.forms.datagridview.cellvaluechanged.aspx
Place your code in the
DataGridView.CellValueChangedevent. Use theSelectedCellsproperty to find the current cell, then check if its value is "Browse From File...". If so, fire off theOpenFileDialog.ShowDialog()call.See this documentation for CellValueChanged: http://msdn.microsoft.com/en-us/library/system.windows.forms.datagridview.cellvaluechanged.aspx