Inserting multiple rows to mysql fetched from while loop - php

Question

I am new to stackoverflow and little bit confused about what to title this question, but let me explain what my problem is.
I am developing survey system where I have option for admin to enter the questions and answers from the back end, so in the front end I am displaying the question, answers using the while loop in the form.
My problem is how to write the code to enter the multiple values (survey filled by the customer) in the mysql. I am attaching my code here.
$(function() {
$('form').bind('submit', function(){
$.ajax({
type: 'post',
url: "/prs/pageCode.php",
data: $("form").serialize(),
success: function() {
alert("Data Saved, Press Next!");
}
});
return false;
});
});
form is below
<form name="form" id="form" class="form-horizontal" action="#"
method="POST">
<fieldset>
<div class="control-group">
<label class="control-label" for="focusedInput">Name</label>
<div class="controls">
<input name="pname" class="input-xlarge focused" id="focusedInput"
type="text" value="" required>
</div>
</div>
<div class="control-group">
<label class="control-label" for="email">Email</label>
<div class="controls">
<input name="pemail" class="input-xlarge focused" id="focusedInput"
type="text" value="" required>
</div>
</div>
<div class="control-group">
<label>We would like to post your comments to internet rating sites,
while we may use your name or an alias name. Please select your
desire:</label>
</div>
<div class="control-group">
<label class="control-label" for="focusedInput"></label>
<div class="controls">
<p>
<label> <input type="radio" name="choice" value="yes" id="choice_0">
Yes, use my name as entered above
</label> <label> <input type="radio" name="choice" value="no"
id="choice_1" onChange="enabletextbox();"> Use the alias name, as
I enter below
</label> <input id="alias" name="alias" type="text"
class="input-xlarge focused">
</p>
</div>
</div>
</fieldset>
</div>
<div class="tab-pane" id="tab2">
<fieldset>
<?php
$i=1;
while($row=mysqli_fetch_array($questions)){
?>
<div class="control-group">
<label class="control-label" for="focusedInput">(<?php echo $i;?>) <?php
$questionid = $row['question_id'];
$question = $row['question'];
echo $row['question']; ?></label>
<div class="controls">
<?php
if($row['answer_type']=="Ratings") {
echo "
<p>
Low<input type='radio' name='rating$i' value='1' id='rating_0'>
<input type='radio' name='rating$i' value='2' id='rating_1'>
<input type='radio' name='rating$i' value='3' id='rating_2'>
<input type='radio' name='rating$i' value='4' id='rating_3'>
<input type='radio' name='rating$i' value='5' id='rating_4'>High
</p>
";
}
else if ($row['answer_type']=="Comments") {
echo "<textarea name='answer' cols='' rows=''></textarea>";
}
$i++;
echo "<br />";
$query2 = "insert into review_details (review_id,survey_id,question_id,question,answer_rating,answer_freeresponse) values (1,$surveyid,$questionid,'$question','$rating$i','$_POST[answer]')";
$result2 = mysqli_query($con,$query2);
if(!$result2) {
echo mysqli_error($result2);
}
?>
</div>
</div>
<?php }?>
</fieldset>
</div>
<div class="tab-pane" id="tab3">
<fieldset>
<div class="control-group">
<label class="control-label" for="focusedInput">Testimonial about
your care by <?php echo $_SESSION['doctorname'];?>
</label>
<div class="controls">
<textarea name="review" cols="5" rows="5" required></textarea>
</div>
</div>
<div class="form-actions">
<input type="submit" class="btn btn-primary"
value="Save & Press Finish"></input>
</div>
</fieldset>
pageCode.php
<?php
session_start();
require_once('config.php');
$con=mysqli_connect(HOST,USER,PASSWORD,DATABASE);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$alias = $_POST['choice'];
if($alias=="yes")
$aliasname = $_POST['pname'];
else
$aliasname = $_POST['alias'];
$sql="INSERT INTO reviews (doctor_id,doctor_name,survey_id,name,email,display_name,overall_review)
VALUES
('$_SESSION[doctorid]','$_SESSION[doctorname]',$_SESSION[surveyid],'$_POST[pname]','$_POST[pemail]','$aliasname','$_POST[review]')";
$result = mysqli_query($con,$sql);
//$result2 = mysqli_query($con,$_SESSION['insertallquery']);
if(!$result) {
echo mysqli_error($result);
}
?>
I actually want to insert the answers into database filled by customers.

Answer 1

You need to store answers in the review table or a separate table its your wish.
but in general, what I understood you have a separate table for questions with the question ids.
So, you need to add column in the review table or a separate table with the question ids as the field names.
and store the respective answers in those column for that user.
For eg. considering adding column in existing reviews table :
INSERT INTO reviews (doctor_id,doctor_name,survey_id,name,email,display_name,overall_review,q1,q2,q3...)
so on according to number of questions you have.
Also, when you add a new question to the questions table you will have to make sure to add a column in this table using ALTER TABLE with the.
For getting the answers as the question are in a while loop, create a array.
and insert those values into mysql using a while loop iterating the array.
The query inside the while loop will be like this :
INSERT INTO reviews (q.$i) VALUES ($answer[$i]) WHERE docter_id='$_SESSION[doctorid]'

Answer 2

You could get the values from IDs/class names or tag names itself then you could store it in an array then you can pass it to your jquery code( $.ajax ) or ($.post) then accept the array variable declared in your javascript
but our tags must contain temp index values example.
Another way is to pass the answers as JSONs then you could use json_encode in php to access the answers