I want code to switch the buttons. If I pressed button1 first time, it must show button2 and vice versa.
<input type="submit" value="asc" name="button1" id="but1">
<input type="submit" value="desc" name="button2" id="but3">
I want code to switch the buttons. If I pressed button1 first time, it must show button2 and vice versa.
<input type="submit" value="asc" name="button1" id="but1">
<input type="submit" value="desc" name="button2" id="but3">
Not sure what you're trying to achieve, but you can use:
$('input[type="submit"]').click(function() {
$(this).hide().siblings('input[type="submit"]').show();
});
You can try the code below:
$('input[type="submit"]').click(function(){
var valueOfButton = $(this).val();
if(valueOfButton == 'asc')
{
$('input[value="asc"]').show();
$('input[value="desc"]').hide();
}
else
{
$('input[value="desc"]').show();
$('input[value="asc"]').hide();
}
});
Try using the following functions:
$(element)click(callback)
will handle the click of the element
$(element).show()
will show the element
$(element).hide()
will hide the element
so a semple code is:
//first hidden the second button
$('#but3').css('display','none')
// handle click of first button
$('#but1').click(function(){
$(this).hide()
$('#but3').show()
});
// handle click of second button
$('#but3').click(function(){
$(this).hide()
$('#but1').show()
});
Here is an example: http://jsfiddle.net/L7zux/1/
One solution without the need for JQuery would be this one:
<input type="button" value="asc" name="button1" id="but1" onClick="document.getElementById('but3').style.display='';this.style.display='none';">
<input type="button" value="desc" name="button2" id="but3" style="display:none;" onClick="document.getElementById('but1').style.display='';this.style.display='none';">
You can also do it this way if you want to use the visibility:
<input type="button" value="asc" name="button1" id="but1" onClick="document.getElementById('but3').style.visibility='visible';this.style.visibility='hidden';">
<input type="button" value="desc" name="button2" id="but3" style="visibility:hidden;" onClick="document.getElementById('but1').style.visibility='visible';this.style.visibility='hidden';">
Using visibility preserves the buttons position. I changed the type from submit to button just out of demonstration reasons.
You can look at both JSFIDDLE demos of these solutions here and here.
document.getElementById('but1').addEventListener('click', function() {
document.getElementById('but1').style.visibility = 'hidden';
document.getElementById('but3').style.visibility = 'visible'; }, false);
document.getElementById('but3').addEventListener('click', function() {
document.getElementById('but3').style.visibility = 'hidden';
document.getElementById('but1').style.visibility = 'visible'; }, false);
If you want to hide button and its placeholder completely, use style.display = 'none'
and style.display = 'block'
. If you put both buttons in div
container with default static
positioning, then both buttons will appear at the same position in container.
Simply Use .toggle()
in jQuery
$('input[type="submit"]').click(function() {
$('input[type="submit"]').toggle();
});
I'm betting your .toggle-radio-switch elements are siblings. Remove .parent() from your code. It isn't needed since .radio-switch-slider is contained directly in .toggle-radio-switch
$(this).find('.radio-switch-slider')
By default when page will load put following code so that your second button will be hide.
$(document).ready(function(e){
$('#but3').hide();
});
After that Put code that were
$('input[type="submit"]').click(function() {
$(this).hide().siblings('input[type="submit"]').show();
});