Question

我有这个方法自动完成名称（字符串），我想创建完成Int Type的其他方法，我尝试但它说不能仅对字符串

使用LIKE作为int类型

 void AutoComplete()
    {
        textBox1.AutoCompleteMode = AutoCompleteMode.Suggest;
        textBox1.AutoCompleteSource = AutoCompleteSource.CustomSource;
        AutoCompleteStringCollection Collection = new AutoCompleteStringCollection();
        con.Open();
        cmd = new SqlCommand("select * from stagiaire",con);
        dr = cmd.ExecuteReader();
        while (dr.Read())
        {
            nom = dr.GetString(1).ToString();
            Collection.Add(nom);
        }

        textBox1.AutoCompleteCustomSource = Collection;
        con.Close();
    }

private void textBox1_TextChanged(object sender, EventArgs e)
    {

        DataView dv = new DataView(dt);
        dv.RowFilter = string.Format("nom like '{0}%'", textBox1.Text);
        dataGridView1.DataSource = dv;
    }
 private void Form1_Load(object sender, EventArgs e)
    {

        da = new SqlDataAdapter("select * from stagiaire",con);
        da.Fill(dt);

    }

Answer 1

对于具有DataType INTEGER的列的

使用如下查询：

select * from Table where Id like 1;

或将其转换为varchar：

 select * from Table where CAST(Id AS VARCHAR(45)) like '1%'

使用Int值自动完成

1 个答案: