更新时间:2023-11-30 19:57:28
也许你的意思是? DEMO
$(#foo)。 $('#checked-a').css(width,40%);
('#checked-a')。 $('#checked-b')。show();
});
else $('#checked-b')。slideUp('fast',function(){
$('#checked-a').css(width,100%);
});
});
});
FIXED
fixed it by editing mplungjan's code to this:
$(function() {
$("#foo").on("click",function() {
if ($(this).is(':checked')) $('#checked-a').show('fast',function() {
$('#checked-b').css("width","60%");
$('#checked-a').css("width","40%");
}) ;
else $('#checked-a').show('fast',function(){
$('#checked-b').css("width","100%").show();
$('#checked-a').css("width","0%").hide();
});
});
});
i would like to have the left div to be 100% (current 60%) when right div (40%) is hidden.
screenshot with checked checkbox:
screenshot with unchecked checkbox:
now i would like to make the left div cover the whole width of the page (100%).
here is my code:
javascript:
<script type="text/javascript">
$(function(){
($('input[name = "foo"]').is(':checked'))
$('#checked-a').slideDown('fast');
});
</script>
div:
<div class="content2" id="checked-a">
<%= yield %>
</div>
html input:
<input type="checkbox" checked="checked" name="foo" value="1" onclick="$(this).is(':checked') && $('#checked-a').slideDown('fast') || $('#checked-a').slideUp('fast');"/>
hope someone could help :)
Perhaps you mean this? DEMO
$(function() {
$("#foo").on("click",function() {
if ($(this).is(':checked')) $('#checked-a').slideDown('fast',function() {
$('#checked-a').css("width","40%");
$('#checked-b').show();
}) ;
else $('#checked-b').slideUp('fast',function(){
$('#checked-a').css("width","100%");
});
});
});