I have been using jQuery 1.8.3, and have two forms and two radio buttons, form will show when click on the radio buttons. what i am doing here is when click on save button, i will check which radio is checked, depends on that form will submit, the problem is after one form submit, that request going as ajax, so i am not refresh the page after one submit, when click on the save button, two times the form will submit, and next time 4 times ...?
js code
$(".save").live("click", function() {
if ($("#b1").is(":checked")) {
$("#lForm1").submit();
}
if ($("#b2").is(":checked")) {
$("#lForm2").submit();
}
});
$('input[type="radio"]').click(function() {
//here toggle the form
});
html
<input type="radio" name="optionsRadios" id="b1" checked>
<input type="radio" name="optionsRadios" id="b2" >
<%= form_tag '/contact/create_check', method: :post, remote: true, id:
'lForm1' %>
<%= form_tag '/contact/create_count', method: :post, remote: true, id:
'lForm2' %>
<input type="button" class="btn btn-success save" value="Save" />
Please do something like this instead
$(function() {
$("#lForm1").on("submit", function(e) {
e.preventDefault(); // stop submission
if ($("#b1").is(":checked")) {
$.ajax({url:$("#lform1").prop("action"),data:$("#lform1").serialize()});
}
else if ($("#b2").is(":checked")) {
$.ajax({url:$("#lform2").prop("action"),data:$("#lform2").serialize()});
}
});
});
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