Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
505 views
in Technique[技术] by (71.8m points)

asp.net - Enable/Disable asp:validators using jquery

I am working with a wizard, where the user can sign up. There is a asp:RadioButtonList with two options, and some of the input fields in the wizard changes when the radiobutton changes. On each field there is some asp:Validators (asp:RequiredFieldValidator for example). The problem is, that when the user submits the page, the validator for the hidden textbox is still popping up.

First, here is the div tags which changes the shown textboxes and the RadioButtonList

<div id="divTxt1">
  <asp:TextBox runat="server" CssClass="text" ID="txtNumber"
       type="number"/>
  <asp:RequiredFieldValidator ID="RequiredFieldValidator1" 
       runat="server" ControlToValidate="txtNumber" EnableClientScript="true" ErrorMessage="Error" ToolTip="Error">*
   </asp:RequiredFieldValidator>
</div>
<div id="divTxt2">
  <asp:TextBox runat="server" CssClass="text" ID="txtNumber2"
       type="number"/>
  <asp:RequiredFieldValidator ID="RequiredFieldValidator2" 
       runat="server" ControlToValidate="txtNumber2" EnableClientScript="true" ErrorMessage="Error2" ToolTip="Error2">*
   </asp:RequiredFieldValidator>
</div>
<div id="radio">
<asp:RadioButtonList ID="RadioButtonList1" runat="server" RepeatDirection="Horizontal">
   <asp:ListItem Value="1" Selected="True">Privat</asp:ListItem>
   <asp:ListItem Value="2">Offentlig</asp:ListItem>
   </asp:RadioButtonList>
</div>

I have tried to solve it using JQuery like the following, which I have read should do the trick, but unfortunately it doesn't:

$(document).ready(function () {

    $('#<%= WizardStep1.ContentTemplateContainer.FindControl("RadioButtonList1").ClientID %> input').change(function () {
        if ($(this).val() == "1") {
            $('#txtNumber').toggle('fast');
            $('#txtNumber2').toggle('fast');     
            ValidatorEnable($('#<%=WizardStep1.ContentTemplateContainer.FindControl("RequiredFieldValidator1").ClientID %>')[0], false);
            ValidatorEnable($('#<%=WizardStep1.ContentTemplateContainer.FindControl("RequiredFieldValidator2").ClientID %>')[0], true);
        }

        if ($(this).val() == "2") {
            $('#txtNumber').toggle('fast');
            $('#txtNumber2').toggle('fast');
            ValidatorEnable($('#<%=WizardStep1.ContentTemplateContainer.FindControl("RequiredFieldValidator2").ClientID %>')[0], false);
            ValidatorEnable($('#<%=WizardStep1.ContentTemplateContainer.FindControl("RequiredFieldValidator1").ClientID %>')[0], true);
        }
    });
});

So, any ideas what's wrong?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I found a better option was to use simply:

document.getElementById("<%=myValidator.ClientID %>").enabled = true;

The ValidatorEnabled option as suggested above automatically calls the validation of the linked control and in my case shows the error message "please enter a value for seller name" which wasn't necessary or desired..

Using the ".enabled = true" option doesn't.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...