【发布时间】:2015-12-15 22:36:30
【问题描述】:
我正在尝试将单选按钮选择的值保存到数据库中,就好像它们没有被动态认定那样保存起来没什么大不了的,但是在我的情况下,比率按钮是在 html 中动态生成的表。
下面我添加了一个html 代码,其中它只有 3 组单选按钮,就像我有 n 个单选按钮生成取决于数据库获取行 s
我想要的是我只想保存提交时选择的单选按钮的值。到Mysql
<table id="attendence_div" width="100%" align="center" cellpadding="2" cellspacing="2" border="0">
<tr align="left" style="background-color:#004080;color:White;">
<td> Student Name</td>
<td>Present</td>
<td>Absent</td>
<td>Leave</td>
</tr>
<tr>
<td>ANITHA S</td>
<td><input type="radio" name="Present0" value="Present"></td>
<td><input type="radio" name="Present0" value="Absent"></td>
<td><input type="radio" name="Present0" value="Leave"></td>
</tr>
<tr>
<td>ANITHA T C</td>
<td><input type="radio" name="Present1" value="Present"></td>
<td><input type="radio" name="Present1" value="Absent"></td>
<td><input type="radio" name="Present1" value="Leave"></td>
</tr>
<tr>
<td>BINDU K V</td>
<td><input type="radio" name="Present2" value="Present"></td>
<td><input type="radio" name="Present2" value="Absent"></td>
<td><input type="radio" name="Present2" value="Leave"></td>
</tr>
</table>
C#代码
protected void Submit(object sender, EventArgs e)
{
string present = Present0.Checked;
string absent = Present1.Checked;
string Leave = Present2.Checked;
string constr = ConfigurationManager.ConnectionStrings["constr"].ConnectionString;
using (MySqlConnection con = new MySqlConnection(constr))
{
using (MySqlCommand cmd = new MySqlCommand("INSERT INTO attendence(absent, present, Leave ) VALUES(@absent, @present,@Leave)"))
{
cmd.Connection = con;
cmd.Parameters.AddWithValue("@absent", absent);
cmd.Parameters.AddWithValue("@present", present);
cmd.Parameters.AddWithValue("@Leave", Leave);
con.Open();
cmd.ExecuteNonQuery();
con.Close();
}
}
}
【问题讨论】:
-
你试过asp radioButtonList控件吗
-
@FatihSert 我是新手,我真的不知道该怎么做:(如果你能帮助我,我会非常感谢你