Retrieving the correct ID.

furrls · December 8, 2010

$qID = '';

$question = 'Question not set';

$answerA = 'unchecked';

$answerB = 'unchecked';

$answerC = 'unchecked';

$answerD = 'unchecked';

$answerE = 'unchecked';

?>

<html>

<head>

function show_alert()

{

alert("Please Click OK to proceed!");

}

</script>

</head>

<table>

<tr>

<td>

<?php

$SQL = "SELECT stu_satisfaction_tblquestions.question_id, stu_satisfaction_tblquestions.question, answer_type.answer1, answer_type.answer2, answer_type.answer3, answer_type.answer4, answer_type.answer5 FROM stu_satisfaction_tblquestions INNER JOIN answer_type ON stu_satisfaction_tblquestions.answers_id=answer_type.answers_id

";

$result = mysql_query($SQL);

while($db_field = mysql_fetch_assoc($result)){

$qID = $db_field['question_id'];

$question = $db_field['question'];

$A = $db_field['answer1'];

$B = $db_field['answer2'];

$C = $db_field['answer3'];

$D = $db_field['answer4'];

$E = $db_field['answer5'];

print $question;

?>

</td>

</tr>

<tr>

<td>

<?php echo $qID;?>

</td>

</tr>

<tr>

<td>

<?php echo $qID;?>

</tr>

</td>

<tr>

<td>

</tr>

</td>

<tr>

<td>

</tr>

</td>

<tr>

<td>

<br>

<?php } mysql_close(); ?>

</tr>

</td>

<tr>

<td>

Please add any comments you may have:</br>

</textarea>

</td>

</tr>

</table>

</form>

</html>

I ran to some problems here. After filling every question up with the answers. When i process the form, i only managed to get QID of 20 and the answer of 1st question as my database have only 20questions. How do i go about in solving, so that i'm able to retrieve all result and update my database correctly

$selected_radio = $_POST['q'];

$qID = $_POST['h'];

echo $selected_radio;

echo $qID;

$SQL = "UPDATE answers SET $selected_radio = $selected_radio + 1 WHERE question_id='$qID'";

$result = mysql_query($SQL);

mysql_close();

print "Thanks for voting!";

?>

THis is my process page, and i echo out the answer to be B and qid as 20.

Buddski · December 8, 2010

Here is a sample of how I would go about doing this.. Your current setup means that every question creates a hidden field with the name 'h' this will not work. You need a HTML array and iterate through it on post.

Example:

// Form Code //
$sql = "SELECT * FROM `q_and_a`";
$sql_query = mysql_query($sql);
while ($data = mysql_fetch_array($sql_query)) {
echo $data['question'].'<br/>';
echo '<input type="radio" name="answers['.$data['question_id']].'" value="A" />';
echo '<input type="radio" name="answers['.$data['question_id']].'" value="B" />';
echo '<input type="radio" name="answers['.$data['question_id']].'" value="C" />';
}

// Post Handler //
foreach ($_POST['answers'] as $qID=>$answer) {
echo 'The person answered '.$answer.' for qID '.$qID.'<br/>';
}

Hope this gives you something to work with.. It havent tested it but its something at least.

Also when posting in future, wrap your code in [ code ] or [ php ] tags

furrls · December 8, 2010

hey, thanks for the codes. but i ran into some syntax errors

Parse error: syntax error, unexpected ']', expecting ',' or ';' in C:\xampp\htdocs\studsurvey.php on line 39

echo '<input type="radio" name="answers['.$data['question_id']].'" value="A" />';

Buddski · December 8, 2010

echo '<input type="radio" name="answers['.$data['question_id'].']" value="A" />';

Whoops.

furrls · December 8, 2010

Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in C:\xampp\htdocs\studsurvey.php on line 36

$SQL = "SELECT stu_satisfaction_tblquestions.question_id, stu_satisfaction_tblquestions.question, answer_type.answer1, answer_type.answer2, answer_type.answer3, answer_type.answer4, answer_type.answer5 FROM stu_satisfaction_tblquestions INNER JOIN answer_type ON stu_satisfaction_tblquestions.answers_id=answer_type.answers_id
";
$result = mysql_query($SQL);
while ($data = mysql_fetch_array($Sresult)) {

furrls · December 8, 2010

oopps, i found the error

furrls · December 8, 2010

as for the post handler codes. Can i do this to update my database the results?

foreach ($_POST['answers'] as $qID=>$answer) {
echo 'The person answered '.$answer.' for qID '.$qID.'<br/>';
$SQL = "UPDATE answers SET $answers = $answers + 1 WHERE question_id='$qID'";
		$result = mysql_query($SQL);
		mysql_close();
		print "Thanks for voting!";
}

my fields in my database is current id,question_id, A, B, C, D, E

Buddski · December 8, 2010

Only one way to find out

But before you try.. Change the variable $answers in your SQL to $answer otherwise you will get undefined errors

furrls · December 8, 2010

ok thanks, if i wanna print out the options for each questions how do i do that? cos currently i have the radio button but didnt display the options or the answers. It is because there are different sets of answers such as - (Satisfied, not satifsied) or (agree,disagree)

should i change it to

echo '<input type="radio" name="answers['.$data['question_id'].']" value="<?php print $A" /><br/>';

?

Buddski · December 8, 2010

Something like this might work.

echo '<input type="radio" name="answers['.$data['question_id'].']" id="'.$data['question_id'].'_'.$A.'" value="'.$A.'" /><label for="'.$data['question_id'].'_'.$A.'">'.$A.'</label><br/>';

I added the ID attribute so you can write clickable labels for the radio buttons

furrls · December 8, 2010

it didnt work, nothing came out. but for $answerA, the word "unchecked" appeared.

<?php
include 'config.php';


$qID = '';
$question = 'Question not set';
//$answerA = 'unchecked';
$answerB = 'unchecked';
$answerC = 'unchecked';
$answerD = 'unchecked';
$answerE = 'unchecked';




?>
<html>
<head>
<script type="text/javascript">
function show_alert()
{
alert("Please Click OK to proceed!");
                  
}
</script>
</head>
<form action="Process1.php" method="POST">
<table>
<tr>
<?php

$SQL = "SELECT stu_satisfaction_tblquestions.question_id, stu_satisfaction_tblquestions.question, answer_type.answer1, answer_type.answer2, answer_type.answer3, answer_type.answer4, answer_type.answer5 FROM stu_satisfaction_tblquestions INNER JOIN answer_type ON stu_satisfaction_tblquestions.answers_id=answer_type.answers_id
";
  
$result = mysql_query($SQL);
while ($data = mysql_fetch_array($result)) {
    $answerA = $db_field['answer1'];
$B = $db_field['answer2'];
$C = $db_field['answer3'];
$D = $db_field['answer4'];
$E = $db_field['answer5'];
echo $data['question'].'<br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" id="'.$data['question_id'].'_'.$answerA.'" value="'.$answerA.'" /><label for="'.$data['question_id'].'_'.$answerA.'">'.$answerA.'</label><br/>';
//echo '<input type="radio" name="answers['.$data['question_id'].']" value="A" /><br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" value="B" /><br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" value="C" /><br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" value="D" /><br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" value="E" /><br/>';
}
?>

Buddski · December 8, 2010

Where is the data for the names stored, the satisfied, unsatisfied stuff. It appears you are storing 'checked' or 'unchecked' in the $answerA variable..

Without a full scope of your system and what is stored where I cant really do much more

furrls · December 8, 2010

those are stored in table called answer_type

answers_id answer1 answer2 answer3 answer4 answer5

1 Strongly Agree Agree Dont Know Disagree Strongly Disagree

furrls · December 8, 2010

anyone?

furrls · December 8, 2010

<?php
include 'config.php';

$SQL = "SELECT stu_satisfaction_tblquestions.question_id, stu_satisfaction_tblquestions.question, answer_type.answer1, answer_type.answer2, answer_type.answer3, answer_type.answer4, answer_type.answer5 FROM stu_satisfaction_tblquestions INNER JOIN answer_type ON stu_satisfaction_tblquestions.answers_id=answer_type.answers_id
";
$result = mysql_query($SQL);
while($db_field = mysql_fetch_assoc($result)){


$A = $db_field['answer1'];
$answerB = 'unchecked';
$answerC = 'unchecked';
$answerD = 'unchecked';
$answerE = 'unchecked';


}


?>

<html>
<head>
<script type="text/javascript">
function show_alert()
{
alert("Please Click OK to proceed!");
                  
}
</script>
</head>
<form action="Process1.php" method="POST">
<table>
<tr>
<?php

$SQL = "SELECT stu_satisfaction_tblquestions.question_id, stu_satisfaction_tblquestions.question, answer_type.answer1, answer_type.answer2, answer_type.answer3, answer_type.answer4, answer_type.answer5 FROM stu_satisfaction_tblquestions INNER JOIN answer_type ON stu_satisfaction_tblquestions.answers_id=answer_type.answers_id
";
  
$result = mysql_query($SQL);
while ($data = mysql_fetch_array($result)) {

echo $data['question'].'<br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" id="'.$data['question_id'].'_'.$A.'" value="'.$A.'" /><label for="'.$data['question_id'].'_'.$A.'">'.$A.'</label><br/>';
//echo '<input type="radio" name="answers['.$data['question_id'].']" value="A"<br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" value="B" /><br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" value="C" /><br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" value="D" /><br/>';
echo '<input type="radio" name="answers['.$data['question_id'].']" value="E" /><br/>';
}
?>
<td>
</tr>
</td>
<tr>
<td>
Please add any comments you may have:</br>
<textarea rows="3" cols="60" name="comments" id="comments">
</textarea>

</td>
</tr>


</table>

<input type="submit" onclick="show_alert()" value="Submit" />
</form>

</html>

I managed to print out the 1st option, but it didnt tally with my database for the question and the set of answers, how do i go about doiing it ?

Anti-Moronic · December 8, 2010

BIG EDIT:

Removed everything I said before.

I have no idea, but when you are selecting lots of columns from the same table you can use *, so in your current statement it would become:

$SQL = "SELECT stu_satisfaction_tblquestions.question_id, stu_satisfaction_tblquestions.question, answer_type.* FROM stu_satisfaction_tblquestions INNER JOIN answer_type ON stu_satisfaction_tblquestions.answers_id=answer_type.answers_id";

One thing which tells me this wouldn't work is your use of answers_id. You shouldn't need to use that and definitely shouldn't have an answers_id in your questions table. You should be using a question_id in your answers table.

Like so:

ON stu_satisfaction_tblquestions.answers_id=answer_type.answers_id

becomes

ON stu_satisfaction_tblquestions.question_id=answer_type.question_id

If that doesn't work you need to seriously rethink how your are structuring your database. Ideally, each answer will be a new row. Not a row for 5 answers, it would be 5 rows for each answer, each with the same question_id.

furrls · December 8, 2010

echo '<input type="radio" name="answers['.$data['question_id'].']" id="'.$data['question_id'].'_'.$A.'" value="'.$A.'" /><label for="'.$data['question_id'].'_'.$A.'">'.$A.'</label><br/>';

so nothing to do with this ? i suppose? just the query right ?

Anti-Moronic · December 8, 2010

read my post again:

"I have no idea".

The rest is a shot in the dark and a very serious suggestion for you to restructure how you are relating questions to answers. It's not right. How can a question have an answer id? The answer is a child of the question. The answer id shouldn't be in the questions table at all. The answers should have question id's and that is how the join works.

Maybe you need to do LEFT JOIN instead.

I'd restructure your database because if I started from scratch right now could have this running very smoothly within 20 minutes. Yet it would take that just to debug what you have. It's simple stuff so it isn't working already there is something fundamentally wrong.

Get your head clear, think about how question_id would work in both tables, with one answer for each row in the answers table. Create your select drop down again based on that.

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Retrieving the correct ID.

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