PHP Sticky For to pass variable to JQuery

I want to pass a php form sticky entry to Jquery upon page refresh.

PHP Code:
	function choose_registration_type() {
		echo '  <h4>I Would Like To:</h4><br />
				<input type="radio" name="team_join" id="create_new_team" value="Create New Team" />Create New Team<br />
				<input type="radio" name="team_join" id="join_existing_team" value="Join Existing Team"'; if(isset($_POST['team_join']) && $_POST['team_join'] == "Join Existing Team") {echo'checked';} echo'/>Join Existing Team<br />
				<input type="radio" name="team_join" id="join_individually" value="Join Individually" />Join Individually';
	}

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So if the form fails, the radio button is still chosen. However, this is within a div that is shown/hidden by JQuery  with a $(document).ready(function()

So since this radio button is already "Checked" I want the div to automatically show.
But, the following does not work:
$(document).ready(function() {
$('input[type="radio"]').checked(function() {
		if ($(this).attr("value") == "Join Existing Team") {
		$("#join_group").show('fast');
		}
	});
});

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LVL 7
rgranlundAsked:
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leakim971Connect With a Mentor PluritechnicianCommented:
What about :
$(document).ready(function() {
        var checked = $('input[type="radio"]:checked');
        var at_least_one_is_checked = checked.length > 0;
        var is_Join_Existing_Team = checked.attr("value") == "Join Existing Team";
	if( checked.length>0 && is_Join_Existing_Team ) {
		$("#join_group").show('fast');
	};
});

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0
 
Alex EneCommented:
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Ray PaseurCommented:
...if the form fails, the radio button is still chosen
What do you mean by if the form fails?
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rgranlundAuthor Commented:
This was exactly what I needed.  I'm slowly learning JQuery and this helped!
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