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 = $row['Option_10'];
?>
<inputstyle="display:none"type="text"class="form-control"value="<?phpecho$qid?>"name="question_<?phpecho$answer; ?>"><?phpfor($j=1;j<=10;$j++)
{
if(isset(${'Option_'.$j}) && !empty(${'Option_'.$j}) )
{
?><div><inputtype="<?phpif($anstype == 'radiobutton'){ echo'radio'; }else{ echo'checkbox'; }?>"name="answer_<?phpecho$answer?>"value="<?phpecho ${'Option_'.$j};?>"id="<?phpecho ${'Option_'.$j}; ?>"><?phpecho ${'Option_'.$j}; ?></div><?php
}
?><br/><?php
}
?><?php
}
}
?>
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