Displaying Data From Database Not Working Properly In PHP/javascript

I have a survey form displaying questions, answertype and number of options from my question table. Here's my code.

Solution 1:

1) Id attribute should be unique for each element.

2) You have much code redundant .reduce the code redundant using for loop .

3) Remove your unwanted display:none style property and jquery . because you can able to handle the empty option in server side code .

PHP : Simplified version of your code

$result = mysqli_query($con,$query);

if($result)
{

    while($row = mysqli_fetch_array($result))
    {

        $questions = $row['questiontitle'];
        $qid = $row['question_id'];
        $anstype = $row['answer_type'];
        $count++;
        $answer++;

        echo "<b>Question ". $count.".)</b> ". $questions. "</br>";


        $option1 = $row['Option_1'];
        $option2 = $row['Option_2'];
        $option3 = $row['Option_3'];
        $option4 = $row['Option_4'];
        $option5 = $row['Option_5'];
        $option6 = $row['Option_6'];
        $option7 = $row['Option_7'];
        $option8 = $row['Option_8'];
        $option9 = $row['Option_9'];
        $option10 = $row['Option_10'];

         ?>


         <input style="display:none" type="text" class="form-control" value="<?php echo $qid ?>" name="question_<?php echo $answer; ?>">

        <?php   
        for($j=1;j<=10;$j++)
        {

             if(isset(${'Option_'.$j}) && !empty(${'Option_'.$j}) )
             { 
             ?>

                   <div>
                        <input type="<?php if($anstype == 'radiobutton'){  echo 'radio'; }else{ echo 'checkbox';  }?>" name="answer_<?php echo $answer ?>" value="<?php echo ${'Option_'.$j};?>" id="<?php echo ${'Option_'.$j}; ?>"><?php echo ${'Option_'.$j};  ?>
                    </div>
             <?php  
             }
             ?>  


            <br/>

        <?php 
        } 
        ?>

                <?php  

    }

}  

?>

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