I want to use PHP variable in jQuery in loop - php

Question

I want to use PHP variable in jQuery.
I am having a 50 students data in table which having textbox and radio button.
By default textbox is hidden but when I clicked on radio button textbox show in table.
$("input[type='radio']").on('change', function() {
if ($(this).val() == "2")
var id = '<?php echo json_encode($absent); ?>';
//$("input[type='text']").show();
else
$("input[type='text']").hide();
});
PHP code:
<input name="nxtmark<?php echo $row[0]; ?>" type="text" id="nxtmark<?php echo $row['id']; ?>" onblur="onleave(this.name);" maxlength="2" placeholder="Enter Mark" hidden="" />
<div class="noerrordiv" id="dnxtmark<?php echo $row[0]; ?>">Mark must not be blank</div>
</td>
<td>
<div align="justify">
<input type="radio" id="rdopresent" name="rdopresent<?php echo $row['id']; ?>" checked="checked" value="1" />Present
<input type="radio" id="rdoabsent" name="rdopresent<?php echo $row['id']; ?>" value="2" />Absent<br />
</div>
</td>
</tr>

Answer 1

You can show / hide text box without using PHP variable. See below code -
$("input[type='radio']").on('change',function() {
//find input text from the previous 'td' for which name starts with 'nxtmark'
var $textBox = $(this).closest('td').prev('td').find('input[name^=nxtmark]');
//show hide text box as per radio button value
if($(this).val() == "2")
$textBox.show();
else
$textBox.hide();
});

Answer 2

If your Problem is just to use PHP variable in jquery then following examples will help you.
if you have to user a php array in jquery then use json_encode in PHP not in SCRIPT.
<?php
$a = array(1,2,3,4);
$a = json_encode($a);
$b = "hello";
?>
<script>
$(document).ready(function(){
alert("<?php echo $a; ?>");
alert("<?php echo $b; ?>");
});
</script>