JavaScript function multiple times

Question

I have a javascript function that when the button is clicked it gets the content of the page, displays the text, and then fades it out. But, when it is clicked a second time, the text doesn't appear.

<script type="text/javascript">
$(document).ready(function(){
$("input[name=buttonup4206]").click(function(){
  $.ajax({url:"vote.php", success:function(result){
    $("#up4206").html(result);
    setTimeout(function(){$("#up4206").fadeOut('slow');},2000);
 }});
});});
</script>
<input type="image" name="buttonup4206" id="buttonup4206" src="vote.png"></input>
<div id="up4206"></div>

Konstantin Dinev · Accepted Answer · 2013-04-26 22:44:24Z

5

Once faded out, your element has display: none, so change this line to:

$("#up4206").css('display', 'block').html(result);

answered Apr 26, 2013 at 22:44

Konstantin Dinev

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3 Comments

Konstantin Dinev Over a year ago

@MarcelGwerder It is pretty much the same thing.

Konstantin Dinev Over a year ago

@elclanrs You got me, I didn't think about that!

Dom Over a year ago

@MarcelGwerder if you want to use display:block use .show(), else use .css('display').

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