Show Next record from the table

[minds_gifts]

hello,

I have multiple-choice questions in a table.
The first question is shown and the user submits the answer.Then i display if the answer is correct or wrong and then i would like to proceed for the next question.
I've done upto checking the answers and now i dont know how to get the next question.

Could somebody pls help me with the snippet.

You can see the working example HERE

Heres the code i got it so far.

<?php

include("contentdb.php");

if ($id == "")
{
$id = "0";
}

$display = mysql_query("SELECT * FROM $table WHERE id > $id ORDER BY id ASC LIMIT 1 ",$db);

if (!$submit) {

echo "<form method=post action=$PHP_SELF>";
echo "<table border=0>";

while ($row = mysql_fetch_array($display)) {

$id = $row["id"];
$question = $row["question"];
$opt1 = $row["opt1"];
$opt2 = $row["opt2"];
$opt3 = $row["opt3"];
$answer = $row["answer"];

echo "<tr><td colspan=3><br><b>$question</b></td></tr>";
echo "<tr><td>$opt1 <input type=radio name=q$id value=\"$opt1\"></td><td>$opt2 <input type=radio name=q$id value=\"$opt2\"></td><td>$opt3 <input type=radio name=q$id value=\"$opt3\"></td></tr>";

}

echo "</table>";
echo "<input type='submit' value='See how you did' name='submit'>";

echo "</form>";

}

elseif ($submit) 

{

echo "</p>";

echo "<p>Here are the answers:";

echo "<table border=0>";
$display = mysql_query("SELECT * FROM $table WHERE id > $id ORDER BY id ASC LIMIT 1",$db);
while ($row = mysql_fetch_array($display)) {

$question = $row["question"];
$answer = $row["answer"];
$q = $row["q"];

echo "<tr><td><br>$question</td></tr>";

if ($$q == $answer) 
        {
        echo "<tr><td>&raquo;you answered ${$q}, which is correct</td></tr>";
        }

elseif ($$q == "") {
echo "<tr><td>&raquo;you didn't select an answer. The answer is $answer</td></tr>";
}

else {
echo "<tr><td>&raquo;you answered ${$q},which is wrong. The answer is $answer</td></tr>";
}

}
echo "</table></p>";

}

?>

[Mordecai]

Are you wanting to display the question on the same page as the "that answer was right/wrong"? If so, you'll have to put another query after it echoes out whether or not the answer was right.
If you want it on a separate page, it's a lot easier. Just echo out a link containing the new $id--the new $id will be the old $id and one ($id+1). The link will look like:
<a href="question.php?id=$id">Next question</a>
You will need to get the $id, however, using $_GET["id"].

[minds_gifts]

Hi Mord,
Thanks for your reply.
I would like to show the second question in a separate page.But how exactly i should do this.could you please rough insert into my code, so i would be to follow the exact flow.

Many thanks

[Mordecai]

In quizpage2.php (I believe that's what you posted the code to), add this line:
$id = $_GET["id"];

Later in the script, after the user is told whether or not his answer is correct, echo out a link to the next question:
$id = $id+1;
echo("<a href=\"quizpage2.php?id=$id\">Next question</a>");

[minds_gifts]

Mordecai,

I still see something troubling me.After the first question is submitted, i click on next question and then when i submit the second question, i see the answers displaying for the first question and then when i click on the next question i get the first question again.
There are only cases possible to print:

if ($$q == $answer) 
		{
		echo "<tr><td>&raquo;you answered ${$q}, which is correct</td></tr>";
		$id = $id+1; 
		echo("<a href=\"quizpage2.php?id=$id\">Next question</a>");
		}

elseif ($$q == "") 
		{
		echo "<tr><td>&raquo;you didn't select an answer. The answer is $answer</td></tr>";
		$id = $id+1; 
		echo("<a href=\"quizpage2.php?id=$id\">Next question</a>");

	}

else {
	echo "<tr><td>&raquo;you answered ${$q},which is wrong. The answer is $answer</td></tr>";
		$id = $id+1; 
		echo("<a href=\"quizpage2.php?id=$id\">Next question</a>");

}

or do you think my query is killing me.

Thanks a lot

[Mordecai]

In the form that you submit the answer in, you'll have to put a hidden value with the ID:
<input type="hidden" name="id" value="$id">

If you're submitting the form via POST, you'l have to make $id = $GET["id"] and $POST["id"].

[minds_gifts]

No Mordecai,

its still the same.when i submit the second question, i still get the answer of first one and then i dont get the 3rd question and so on.

Any other ideas??

Thank you so much

[minds_gifts]

hello,

would somebody tell me where would be the error be??

Many thanks

[kevinsequeira]

a few questions ..

after you submit the 1st question, u display wheter answer is correct / wrong and then show the 2nd question ,answer that and then move to the next question.

1. are the questions stored in the database sequentially ? ie are the id for subsequent questions 1,2,3,4,5 ...

2. say if you're on question 5 , u dont care about the earlier answers for questions 1,2,3 right ??

and finally post ure edited / updated code so that i can try to help u out

reg
kevin

[kevinsequeira]

try this

<?php
include("contentdb.php");
if (!isSet($_POST['id'])){
	$id = 1;
}else{
	$id = $_POST['id'];
}

if(isSet($_POST['submit'])){
     $answer = $_POST['answer'];
     $check = mysql_query("SELECT * FROM $table WHERE id = $id ",$db);
     $row = mysql_fetch_array($check);
     $question = $row["question"];	 
     if($answer == ""){
          $output = "No answer selected";
     }elseif($row["answer"]!=$answer){
     	 $output = "wrong answer selected";	
     }else($row["answer"] == $answer){
     	 $output = "Correct answer";	
     } 
?>
    <table border=0>
	<tr>
     	<td>Previous question was <?php print ucwords($question);?></td>
	</tr>
	<tr>
     	<td><?php print ucwords($output);?></td>
	</tr>	
	</table>
<?php 
}

$id = $next_id;
$next_id = $id + 1;

$display = mysql_query("SELECT * FROM $table WHERE id = $id",$db);
 ?>
    <form method="post" action="<?php print($PHP_SELF);?>">
    <table border=0>
<?php
    $row = mysql_fetch_array($display);
    $id = $row["id"];
    $question = $row["question"];
    $opt1 = $row["opt1"];
    $opt2 = $row["opt2"];
    $opt3 = $row["opt3"];
?>			 
    <tr><td colspan=3><br><b><?php print ucwords($question);?></b></td></tr>
    <tr>
     <td><?php print($opt1); ?><input type=radio name="answer" value="<?php print($opt1); ?>"></td>
     <td><?php print($opt2); ?><input type=radio name="answer" value="<?php print($opt2); ?>"></td>
     <td><?php print($opt3); ?><input type=radio name="answer" value="<?php print($opt3); ?>"></td>
</tr>

</table>
<input type="submit" value="See how you did" name="submit">
<input type="hidden" value="<?php print($id); ?>" name="id">
<input type="hidden" value="<?php print($next_id); ?>" name="next_id">		
</form>

reg
kevin

[minds_gifts]

Hi kevin,

Many thanks for modifying my code.

I get the Parse error: parse error, unexpected '{' in ........

on line No

and this is what I've on that line:

else($row["answer"] == $answer)
	 {
          $output = "Correct answer";    

     } 

Thanks for your time.

[kevinsequeira]

sorry my mistake change }else($row["answer"] == $answer){
to
}elseif($row["answer"] == $answer){
reg
kevin

[minds_gifts]

Thanks for the quick reply kevin.
but, now i get the error here:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ......
Here are the lines which it point out:

$row = mysql_fetch_array($display);
    $id = $row["id"];
    $question = $row["question"];
    $opt1 = $row["opt1"];
    $opt2 = $row["opt2"];
    $opt3 = $row["opt3"];

I've also one more question regarding the previous error - you used if and elseif and also elseif.
Normally, the normally is if(condition) elseif(condition) else.
Can we also mention the statements as you said.

Thanks once again

[kevinsequeira]

try this
$query ="SELECT * FROM $table WHERE id =". $id;
echo $query;
$display = mysql_query($query,$db);
if(mysql_error()){
echo mysql_error();
}
if(mysql_num_rows($display) > 0){
echo "rows found";
}else{
echo "no rows found";
}

also try running the $query directly in your database to see if you are getting any results.
im assuming that u have values for $table and have succesfully connected to the database.
also do u have a record where id is 1 because thats what it is set to initially

reg
kevin
what is the output now

[kevinsequeira]

if your ids are not sequential and all present like
1
2
3
4
5

then this may not work because if its like this
1
3
4
5

at one stage id in the code will be 2 and it wont get a result for it.
if thats the case ure better of changing the queries to
SELECT * FROM $table WHERE id > $id ORDER BY id ASC LIMIT 1
like before

reg
kevin

[minds_gifts]

Kevin,

The id's are sequential. I'm positive- my id's are: 1, 2, 3, 4, 5.

[kevinsequeira]

ok, so what was the output after the last changes, the ones where you try to print out the query - my suggestion 3 posts earlier

reg
kevin

[minds_gifts]

I executed the query directly in phpmyadmin.I got the first row displayed, i.e with id 1.

Working example HERE

Thanks a lot

[kevinsequeira]

😕 :mad: :mad: :mad:
my mistake, pls make this change

if (!isSet($_POST['id'])){

$id = 1;
 $next_id = 1;

}else{

$id = $_POST['id'];
$next_id = $_POST['next_id'];

}

also print the query, $id and $next_id on the screen to debug

reg
kevin

[minds_gifts]

Thanks kevin, its working now:-)

But, how shld i modify the code in-order to show only the answer after each question is submitted and then provide the link for the next question.
Also, after the last question is reached, i dont want to show any link.
This is what i need to do actually.

Thanks once again,