Question

我需要用MS-Access中的链接打开一些文件夹。我的代码如下，但不起作用：

        DataSet DSOne = new DataSet();
        OleDbDataAdapter adp = new OleDbDataAdapter();
 private void button2_Click(object sender, EventArgs e)
        {

            adp.SelectCommand = new OleDbCommand();
            adp.SelectCommand.Connection = oleDbConnection1;
            adp.SelectCommand.CommandText = " select Folder from Sheet where ID  like '%" + textBox9.Text + "%' or OfficialDossier like '%" + textBox9.Text + "%' or Family like '%" + textBox9.Text + "%' or FirstName like '%" + textBox9.Text + "%' or approve like '%" + textBox9.Text + "%' or City like '%" + textBox9.Text + "%' or Department like '%" + textBox9.Text + "%' or Organization like '%" + textBox9.Text + "%' ";
            adp.SelectCommand.Parameters.AddWithValue("@ID", textBox9.Text);
            adp.Fill(DSOne, "Folder");
            string p = "Folder";
            string args = string.Format("/e, /select, \"{0}\"", p);
            ProcessStartInfo info = new ProcessStartInfo();
            info.FileName = "explorer";
            info.Arguments = args;
            Process.Start(info);     
        }

此代码分为两部分：一部分直到adp.Fill(SDOne, "Folder");并且它正常工作！我得到DataGridView datasource和datamember的结果。 string p = "Folder";的第二部分也起作用了！当我获取字符串地址时，此代码打开它。我认为问题是在adp.Fill (DSOne, "Folder");和String之间转换，我不知道如何将datamember放到字符串中。谢谢你的帮助！

Answer 1

您似乎执行命令从数据库中检索文件夹（稍后会详细介绍）但您忘记读取该值并使用它来打开文件系统文件夹

adp.Fill(DSOne, "Folder");
if(DSOne.Tables[0].Rows.Count > 0)
{
    // Here you are sure that the first table of the dataset contains at least one record.
    // So you could retrieve it (Rows[0]) and then access the first column 
    // of that row (Rows[0]["Folder"])
    string p = DSOne.Tables[0].Rows[0]["Folder"].ToString();
    string args = string.Format("/e, /select, \"{0}\"", p);
    ProcessStartInfo info = new ProcessStartInfo();
    info.FileName = "explorer";
    info.Arguments = args;
    Process.Start(info);     
}

构建sql命令的任何类型的字符串连接都注定要失败。您不应该使用该方法，但始终使用参数化查询。在上面的代码中，似乎您添加了一个参数，并且从不使用此参数。我建议将代码的这一部分改为

adp.SelectCommand = new OleDbCommand();
adp.SelectCommand.Connection = oleDbConnection1;
adp.SelectCommand.CommandText = @"SELECT Folder FROM Sheet 
                    WHERE ID LIKE ? or OfficialDossier LIKE ? or Family LIKE ? or 
                    FirstName LIKE ? or approve LIKE ? or City LIKE ? or 
                    Department LIKE ? or Organization LIKE ?";
string parameterValue = "%" + textBox9.Text + "%";
adp.SelectCommand.Parameters.AddWithValue("@p1", parameterValue);
adp.SelectCommand.Parameters.AddWithValue("@p2", parameterValue);
adp.SelectCommand.Parameters.AddWithValue("@p3", parameterValue);
adp.SelectCommand.Parameters.AddWithValue("@p4", parameterValue);
adp.SelectCommand.Parameters.AddWithValue("@p5", parameterValue);
adp.SelectCommand.Parameters.AddWithValue("@p6", parameterValue);
adp.SelectCommand.Parameters.AddWithValue("@p7", parameterValue);
adp.SelectCommand.Parameters.AddWithValue("@p8", parameterValue);

OleDb需要添加到命令集合中的确切参数数量，因为查询中存在多个占位符且顺序相同。在您的情况下，您始终使用由文本框提取的相同值并与sql通配符连接。因此，添加8个具有相同值的参数似乎是正确的。

如何在C＃.net中打开文件夹

1 个答案: