checkbox value displaying twice - php

Question

I am using jquery to get the value of a checkbox. However, what is happening is that the value is getting duplicated and seems to be getting values for all checkboxes in the while loop. I would be grateful if someone could point out my error. Thank you.
UPDATE: Current code. Now only selecting first entry. No output on further checkbox clicks.
PHP Code
while($row = mysql_fetch_array($result))
{
$ticket = $row['ticket_frm'];
$rowdate = date("d/m/Y",strtotime($row['date_frm']));
$id = $row['id_frm'];
$from = $row['from_frm'];
$subject = $row['subject_frm'];
$message = $row['message_frm'];
$myString = <<<EOF
<span><input id="check" type="checkbox" name="delete" value="<?php echo $ticket ?>"></span>
<div class='msgTrue buttonMailTrue' data-message='%s' data-subject='%s' data-rowdate='%s' data-from='%s'>
<img src="images/sml_new_mail_icon.gif" class="mailIcon" alt="" />$subject;
<div class="rowdate">$rowdate</div><br />
<span class="mailFrom">$from</span>
</div>
<p class="checked"></p>
<!-- The following end tag need to be at the start of the line -->
EOF;
printf($myString, $message, $subject, $rowdate, $from);
} echo '<p class="checked">'.'</p>';
jQuery Code
$(function() {
$("#check").click(function() {
var isChecked = $(this).prop("checked"); // or $(this).prop("checked")
if (isChecked) {
$("p.checked").html("Checkbox is checked: <b>True</b>");
} else {
$("p.checked").html("Checkbox is checked: <b>False</b>");
}
});
});

Answer 1

According your code and selectors, there are many input class="check" elements on your page.
You find them by class name, so this is why they are duplicate.
Use id-attribute and $("#id") syntax to get right values, or use this keyword in your code:
//var isChecked = $('.check').is(':checked'); - wrong
var isChecked = $(this).is(':checked'); // right
Update:
You didn't provide unique id for your elements. Of cause this doesn't work. Your error is the same as earlier.
Try this code:
<input id="check$ticket" type="checkbox" name="delete" value="<?php echo $ticket ?>">
Same for javascript - you have to find element by it's unique id
$("#check" + TICKET_NUMBER_HERE).click(function() {
var isChecked = $(this).prop("checked"); // or $(this).prop("checked")
if (isChecked) {
$("p.checked").html("Checkbox is checked: <b>True</b>");
} else {
$("p.checked").html("Checkbox is checked: <b>False</b>");

Answer 2

You should use this to work with current checkbox:
$(".check").click(function() {
var isChecked = this.checked; // or $(this).prop("checked")
if (isChecked) {
$("p.checked").html("Checkbox is checked: <b>True</b>");
} else {
$("p.checked").html("Checkbox is checked: <b>False</b>");
}
});

Answer 3

Change your jquery code to. As $('.check') returns array of object it is by default checking property of 1st object only
$(function() {
$('.check').click(function(){
var isChecked = $(this).is(':checked');
alert(isChecked);
if(isChecked)
$('p.checked').html('Checkbox is checked: <b>True</b>');
else
$('p.checked').html('Checkbox is checked: <b>False</b>');
});
});

Answer 4

Try
$(".check").click( function() {
$("p.checked").html(
"Checkbox is checked: <b>" + $(this).is(":checked") + "</b>" );
}