通过在经典asp中按Enter键，onChange事件无法正常工作

问题描述：

在下面的代码片段中，如果我在文本框中输入详细信息后按输入，则onChange事件无法正常工作。但是如果我在文本框中输入详细信息后单击页面，则事件正常。

s = s&< input type =text name =xyzsize =1önChange=return abc（this.form，1）value =&pageno&style =border：1px solid＃C0C0C0; padding：1 ; font-family：Arial; font-size：8pt>

谢谢

In the below code snippet onChange event is not working if I am pressing 'enter' after putting details in the text box. But the event is working correctly if I m clicking on the page after entering the details in the textbox.

s=s&"<input type=""text"" name=""xyz"" size=""1"" önChange=""return abc(this.form,1)"" value=" & pageno &" style=""border: 1px solid #C0C0C0; padding: 1; font-family:Arial; font-size:8pt"">

Thanks

答

单击enter并使用JavaScript回发时使用EnterEvent来捕获事件，该回发使用按钮的id值将你带到cs页面

< asp： textbox id =TextBox1clientidmode =Staticrunat =serveronkeypress =return EnterEvent（event）xmlns：asp =＃unknown>

< asp：button id =Button1runat =serverstyle =display：nonetext =Buttonxmlns：asp =＃unknown>

Use EnterEvent to capture the event when enter is clicked and using JavaScript postback which uses the id value of the button which ll take u to the cs page

<asp:textbox id="TextBox1" clientidmode="Static" runat="server" onkeypress="return EnterEvent(event)" xmlns:asp="#unknown">

<asp:button id="Button1" runat="server" style="display:none" text="Button" xmlns:asp="#unknown">

function EnterEvent(e) {
        if (e.keyCode == 13) {
            __doPostBack('<%=Button1.UniqueID%>', "");
        }
    }

protected void Button1_Click(object sender, EventArgs e)
    {

    }

谢谢Arjun，

我以不同的方式使用了你的建议而且它有效。我使用onchange和onkeypressed事件如下：

s = s&< input type =textname =xyzsize = 1önChange=返回abc（this.form，1）onkeypress =xyz（e）>

函数xyz（a ，e，1）

{

if（e.keyCode == 13）

{

abc（ a，1）;

}

}

Thanks Arjun,
I used your suggestion in a different way and it worked. I used both onchange and onkeypressed event as below:

s=s&"<input type=""text"" name=""xyz"" size=""1"" önChange=""return abc(this.form,1)"" onkeypress=""xyz(e)"">

function xyz(a,e,1)
{
if(e.keyCode==13)
{
abc(a,1);
}
}

通过在经典asp中按Enter键，onChange事件无法正常工作

相关推荐