I'm using jquery to show/hide a form text input depending on the value of a checkbox:

Code:
$(document).ready(function(){
 
  $("input[name$='usr_type']").click(function(){
  var radio_value = $(this).val();
    if(radio_value=='1') {
      $("#typeother").show("slow");
    }
    else if(radio_value=='2') {
      $("#typeother").hide();
     }
  });
  $("#typeother").hide();
});
Above works great.

However, I also need to SHOW the form input if a returned database value matches 1 as well.

PHP Code: [Select]
Code:
if ($gtprsn['usr_type'] == 1); {  ... show the freaking text input ... }

Nothing I've tried so far works. I can add an if statement to the input itself to add a style and it will display:

PHP Code: [Select]
Code:
<?php if ($gtprsn['usr_type'] == 3): ?>style="display:block !important;"<?php endif; ?>

But this then overrides the jquery and if I check the other radio button (value 2), the input still shows...

Suggestions?