Why isn't the PHP if condition running within my Jquery?

Question

I have the following ajax code:

<script type='text/javascript'>
$(document).ready(function () {
    $('.questionform').on('submit', function(e) {
        e.preventDefault();
        $.ajax({  
            url :  "aplaygroundajaxtest.php",
            type: "POST",
            data: $(this).serialize(),
            success: function (data) {     
 '<?php
if($data3["{$QuestionType}Percent"]<100){ ?>' alert("Alert Message OnClick"); '<?php ;} ?>'
            },
        });
    });
});
</script>

If the value in column {$QuestionType}Percent is less than 100, I want it to alert a message. However, it does not work. the source code shows the following:

<script type='text/javascript'>
$(document).ready(function () {
    $('.questionform').on('submit', function(e) {
        e.preventDefault();
        $.ajax({  
            url :  "aplaygroundajaxtest.php",
            type: "POST",
            data: $(this).serialize(),
            success: function (data) {     
 '' alert("Alert Message OnClick"); ''
            },
        });
    });
});
</script> 

I've tried escaping the " with \ and I've also tried double quotes around the php (in which case, the PHP still disappears in the source code, leaving empty double quotes).

I know that the issue does not lie with $data3["{$QuestionType}Percent"]<100 as it functions correctly elsewhere in my code - and the value in the column is indeed less than 100.

EDIT / UPDATE ---------------------------------------------->

The issue was two-fold: 1) I needed to get rid of the single quotes around PHP tags; 2) it wasn't receiving the value of $data3["{$QuestionType}Percent correctly.

In the URL to which the form posts (aplaygroundajaxtest.php), I corrected the latter issue; now, I'm using the variable $QuestionTypePercent. In the code below, print_r($QuestionTypePercent); shows 14 in the source and the if condition is correctly interpreted. The source shows 14 alert("Alert Message OnClick"); and I get the error Uncaught SyntaxError: Unexpected identifier. When I remove print_r($QuestionTypePercent); the error goes away and the code works. Why is print_r($QuestionTypePercent); interfering with my alert in the code below?

<script type='text/javascript'>
$(document).ready(function () {
    $('.questionform').on('submit', function(e) {
        e.preventDefault();
        $.ajax({  
            url :  "aplaygroundajaxtest.php",
            type: "POST",
            data: $(this).serialize(),
            success: function (data) {        
  <?php
print_r($QuestionTypePercent); // RETURNS 14 CURRENTLY
          if($QuestionTypePercent < 100){?>
                alert("Alert Message OnClick"); //REDIRECTSME TO aplaygroundajaxtest.php
                <?php } ?>

            },
        });
    });
});
</script>

Answer 1

If you want to remove the single-quotes, then remove the single-quotes. (You also have an errant semi-colon to remove.) Instead of this:

'<?php if($data3["{$QuestionType}Percent"]<100){ ?>' alert("Alert Message OnClick"); '<?php ;} ?>'

You want this:

<?php if($data3["{$QuestionType}Percent"]<100){ ?> alert("Alert Message OnClick"); <?php } ?>

The PHP code is functioning just fine, as you can clearly see in your output. But outside of the PHP code you have random single-quotes that don't belong there. PHP never touches those, it just outputs whatever's on the page that you put there.

Answer 2

Remove the

'

This should work

<script type='text/javascript'>
    $(document).ready(function () {
        $('.questionform').on('submit', function (e) {
            e.preventDefault();
            $.ajax({
                url: "aplaygroundajaxtest.php",
                type: "POST",
                data: $(this).serialize(),
                success: function (data) {
                    <?php
                    if($data3["{$QuestionType}Percent"] < 100){ ?>
                    alert("Alert Message OnClick");
                    <?php } ?>
                }
            });
        });
    });
</script>

Answer 3

I have suggestions for you,

First, separate the PHP from your jquery not to make it messy, like this:

<?php
$ques = 0;
if($data3["{$QuestionType}Percent"]<100){ 
  $ques = 1;
}
?>  
<input type="hidden" value="<?php echo $ques; ?>" id="questype"/>
<script type='text/javascript'>
$(document).ready(function () {
$('.questionform').on('submit', function(e) {
    $.ajax({  
        url :  "aplaygroundajaxtest.php",
        type: "POST",
        data: $(this).serialize(),
        success: function (data) {      
           if( document.getElementById('questype').value == 1){
                alert("Alert Message OnClick");
           }
        }
    });
    e.preventDefault();        
   });
  });
 </script>

Or else, just echo out the alert if the if statement is true.

    $.ajax({  
        url :  "aplaygroundajaxtest.php",
        type: "POST",
        data: $(this).serialize(),
        success: function (data) {     
          <?php
           echo "if($data3['{$QuestionType}']Percent < 100 ){ 
            alert('Alert Message OnClick'); 
           }"; 
          ?>
        }
    });

Take note that this file is .php, therefore prior code is the php.