Question

似乎使用Like需要一些确切的格式，我不知道。这是代码：

private void DoFilter()
{
    using (connection = new SqlConnection(connectionString))
    using (SqlDataAdapter adapter = new SqlDataAdapter("SELECT * FROM Customers WHERE CompanyName = @filter + '%'", connection))
    {
        adapter.SelectCommand.Parameters.AddWithValue("@filter", txtFilter.Text.Trim());

        DataTable TCustomers = new DataTable();
        adapter.Fill(TCustomers);

        lstCustomers.DisplayMember = "CompanyName";
        lstCustomers.ValueMember = "Id";
        lstCustomers.DataSource = TCustomers;

    }
}

该代码未引发任何错误。它只是使ListBox（lstCustomers）为空。（txtFilter是一个TextBox，其中包含过滤器字符串）。

Answer 1

使用like operator代替=

SELECT * FROM Customers WHERE CompanyName like @filter + '%'"

Answer 2

您没有使用LIKE运算符，而是使用了=运算符。

应为：

using (SqlDataAdapter adapter = new SqlDataAdapter(
    "SELECT * FROM Customers WHERE CompanyName LIKE @filter + '%'", connection))

SQL的编写方式与末尾的%完全匹配，因此不会得到任何结果。

Answer 3

“赞”应用作（SELECT * FROM Customer WHERE CompanyName LIKE @filter +'％'）：

private void DoFilter()
{
    using (connection = new SqlConnection(connectionString))
    using (SqlDataAdapter adapter = new SqlDataAdapter("SELECT * FROM Customers WHERE CompanyName LIKE @filter + '%'", connection))
    {
        adapter.SelectCommand.Parameters.AddWithValue("@filter", txtFilter.Text.Trim());

        DataTable TCustomers = new DataTable();
        adapter.Fill(TCustomers);

        lstCustomers.DisplayMember = "CompanyName";
        lstCustomers.ValueMember = "Id";
        lstCustomers.DataSource = TCustomers;

    }
}

SQL LIKE表达式给出空结果

3 个答案: