use*_*403 8 c# datagridview int32 winforms
我有一个表单,当我从ComboBox中选择一个列名,并在文本框中键入它时,它会过滤并在DataGridView中显示搜索的条件.当我搜索"引用"时,它是一个int数据类型,也是标识和主键.我收到错误消息:
"无法在System.Int32和System.String上执行'Like'操作."
我的代码是
DataTable dt;
private void searchForm_Load(object sender, EventArgs e)
{
SqlCeConnection con = new SqlCeConnection(@"Data Source=|DataDirectory|\LWADataBase.sdf;");
SqlCeDataAdapter sda = new SqlCeDataAdapter("select * from customersTBL", con);
dt = new DataTable();
sda.Fill(dt);
dataGridView1.DataSource = dt;
comboSearch.Items.Add("[Reference]");
comboSearch.Items.Add("[First Name]");
comboSearch.Items.Add("[Surename]");
comboSearch.Items.Add("[Address Line 1]");
comboSearch.Items.Add("[Address Line 2]");
comboSearch.Items.Add("[County]");
comboSearch.Items.Add("[Post Code]");
comboSearch.Items.Add("[Contact Number]");
comboSearch.Items.Add("[Email Address]");
}
private void searchTxt_TextChanged(object sender, EventArgs e)
{
if (comboSearch.SelectedItem == null)
{
searchTxt.ReadOnly = true;
MessageBox.Show("Please select a search criteria");
}
else
{
searchTxt.ReadOnly = false;
DataView dv = new DataView(dt);
dv.RowFilter = "" + comboSearch.Text.Trim() + "like '%" + searchTxt.Text.Trim() + "%'";
dataGridView1.DataSource = dv;
}
}
Run Code Online (Sandbox Code Playgroud)
Gra*_*ICA 11
将数字转换为过滤器内的字符串:
dv.RowFilter = string.Format("CONVERT({0}, System.String) like '%{1}%'",
comboSearch.Text.Trim(), searchTxt.Text.Trim());
Run Code Online (Sandbox Code Playgroud)
归档时间: |
|
查看次数: |
10344 次 |
最近记录: |