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webdeveloper123

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Posts posted by webdeveloper123

    Stuck on my next step for quiz app

    in Application Design

    Posted · Edited by webdeveloper123

    18 hours ago, Strider64 said:

    though I have a Trivia game that is what you are looking for? 

    That's a really good trivia game Strider. I want something very similar to that.

    18 hours ago, kicken said:

    That removes the complexity of dynamically loading the questions.

    If I did want to load them dynamically, would I be able to reuse the below code? 

     <?php

include 'includes/db.php';

//Assume no results.
$results = [];
$questionId = $_GET['question'] ?? '';
if ($questionId){
    //Only run the query if we have an id.
    $sql = "SELECT q.question, o.choice, q.image 
        FROM questions q 
        JOIN quiz_options o 
          ON q.qid = o.qid 
        WHERE q.qid = :id;";
    $stmt = $pdo->prepare($sql);
    $stmt->execute(['id' => $questionId]);

    //Fetch all the result rows
    $results = $stmt->fetchAll(PDO::FETCH_NUM);
}

//If we have any results
if ($results){
    //Display them
    $last_q = 'x';    // a non-existent value
    $choice_num = 0;
    foreach ($results as [$q, $c, $i]){
        if ($q <> $last_q) // starting a new question so close the last div and begin a new one
        {
            if ($last_q <> 'x') // is this the first question? If not, close the last question's box
            {
                echo "</div>";
            }
            echo "<div class='quizContent'>";  // start the new question box
            echo "<p>$q</p>"; // show the question  - figure out later how to prevent repetition (which I did earlier)

            $pathx = "images/db_images/";

            echo '<img class="picture" src="' . $pathx . $i . '">';

            $choice_num = 0;  // new question; new choice count
        }

        $choice_class = 'choice' . $choice_num; // setup the appropriate choice class
        $last_q = $q;   // remember the current question
        //echo "<p>$q</p>"; // show the question  - figure out later how to prevent repetition (which I did earlier)
        echo "<div class='$choice_class'>";   // a box for this choice
        echo "<p>$c</p>";
        echo "</div>";    // close this choice's box
        $choice_num++;    // bump the choice cnt
    }
    echo "</div>";    // close the last question box
} else {
    //Display an error
    $errorAnswer = 'Question does not exist';
}

?>


<?php 
if (isset($errorAnswer)) {
     
    echo "<p class='quizError'>$errorAnswer</p>";
    
}
?>

    Or would it be a matter of writing new code that generates a HTML document that looks like the html in your fiddle?

    18 hours ago, kicken said:

    step 1 is just getting your JavaScript sorted out for marking your correct/incorrect answers and showing the next button

    Yes, I thought the next step was the "Next" button and I was going to do something like create a button, put 1 in there related to question number 1 and then increment it inside the link go to the question 2.

    But I had a bad feeling that if I did that it would just be a waste of time because of all the JS involved

    Thanks kicken

    Stuck on my next step for quiz app

    in Application Design

    Posted

    Hey guys,

    I've been developing this app for over a month now and I just sort of built it as I went along with no real design or specifications (apart from an ERD). I knew what I wanted but sort of used an ad-hoc manner in development. Now I have got to a stage where I'm not sure what my next step is.

    The app I am building is a quiz, loosely based on this quiz: https://www.bbc.co.uk/news/world-64825206

    What I want is when the URL is loaded, to show the quiz Container with a "Start" Button. So a blank grey box with a "Start" Button. When the start button is pressed, then the first question comes up. Then if you visit the link above you will understand the next part. 

    You see there is 3 choices, but mine will be 4 and you get the DOM scripting effect when You choose the correct answer and it goes green. Then you get green tick with a "Correct!". But I don't want the extra text underneath  to explain why the answer was right (or wrong) 

    But if you choose the wrong answer, you get a red X with "Wrong!" and then the correct answer goes green. But I don't want the percentages of how many people chose the choice. And then if you get the answer right or wrong, it will display a "Next" button to move onto the next question. I want the Next button done in AJAX so it loads in the quiz container without reloading the page. 

    Then you get 1 point for one correct answer and at the end of the quiz you get your total amount based on how many you got right. 

    That's about it. 

    My home page looks like this at the moment: (e.g www.domain.com/index.php) 

    homepage.thumb.png.32c74813d1dd00215eb82130fe14ff5b.png

    Now obviously it says "Question does not exist" because I changed the quiz to use GET to scroll through the questions. obviously if I add  

    ?question=1

    to the end of the URL it will display question 1.

    Here is what the questions look like:

    quiz_ss.thumb.jpg.cc3e05287545476d67540e898027702e.jpg

     

    Sorry for the long post but I'm really stuck on my next move. I have a bad feeling I've been developing the app the wrong way round.

    Many thanks

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted

    19 minutes ago, ginerjm said:

    In your second block of code you are showing us something BUT WHERE IS IT LOCATED IN THE FIRST BLOCK???

     

    Oh I think I know what you mean. Your referring to: 

    
  if (!$results) {
   
 $errorAnswer = 'That Id was not found';
} else {
    $errorAnswer = '';
}

    Well, it's right there in the first block, You wrote the code: 

    while(list($q, $c, $i) = $results->fetch(PDO::FETCH_NUM))

     

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted

    3 minutes ago, ginerjm said:

    You're checking if questionid exists but then saying 'no matching id' instead of 'Id is required'.  That s/b changed

     

    Ok, a minor grammar edit

    6 minutes ago, ginerjm said:

    Then you continue to run the script and try to work with the query results that YOU NEVER RAN.  That s/b changed.

     

    are you referring to $questionId?

    7 minutes ago, ginerjm said:

    In your second block of code you are showing us something BUT WHERE IS IT LOCATED IN THE FIRST BLOCK???

     

    I don't understand. Are you referring to the if statement that checks for NULL

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted · Edited by webdeveloper123

    Hey,

    Got a minor problem. I edited your code, firstly to add images and secondly to replace the PK in the WHERE clause to a placeholder which grabs the value using GET and then to display the correct question. E.g (www.domain.com/index.php?question=5) to display question 5. All is well and it's working. Then I did some error handling such that if the question number is not in the URL (e.g www.domain.com/index.php?question= ) then it will echo an error message , which is also fine.

    Now I am trying to add an error message so that if the user edits the url and puts for example: 20 (which does not exist as there are only 15 questions (1-15) then I want an error message. Again this works, but the same error also shows for a valid question id. Here is the revised code: 

      <?php

include 'includes/db.php';

$questionId = $_GET['question'] ?? '';
$idError = '';

$sql = "SELECT q.question, o.choice, q.image 
    FROM questions q 
    JOIN quiz_options o 
      ON q.qid = o.qid 
    WHERE q.qid = :id;";
$results = $pdo->prepare($sql);



if ($questionId) {
    $results->execute(['id' => $questionId]);
   
} else {
    $idError = 'No matching id';
}



// this produces rows that contain a question and a choice for that question.
// That means you have to handle the multiple choices by NOT showing the same question over and over.

$last_q = 'x';    // a non-existent value
$choice_num = 0;
while(list($q, $c, $i) = $results->fetch(PDO::FETCH_NUM))
    // retrieve the contents of a single row as 2 vars
{
  if($q <> $last_q) // starting a new question so close the last div and begin a new one
  {
    if ($last_q <> 'x') // is this the first question? If not, close the last question's box
      echo "</div>";
    echo "<div class='quizContent'>";  // start the new question box
        echo "<p>$q</p>"; // show the question  - figure out later how to prevent repetition (which I did earlier)

         
         $pathx = "images/db_images/";
       
           
        echo '<img class="picture" src="'.$pathx.$i.'">';
         

    $choice_num = 0;  // new question; new choice count


  }


  $choice_class = 'choice' . $choice_num; // setup the appropriate choice class
  $last_q = $q;   // remember the current question
  //echo "<p>$q</p>"; // show the question  - figure out later how to prevent repetition (which I did earlier)
  echo "<div class='$choice_class'>";   // a box for this choice
   echo "<p>$c</p>"; 
  echo "</div>";    // close this choice's box
  $choice_num++;    // bump the choice cnt


}
echo "</div>";    // close the last question box
if ($q === NUll && $c === NULL && $i === NULL) {
     $errorAnswer = 'Question does not exist';
 } else {
    $errorAnswer = '';
}


?>


<?= $idError ?>
<?= $errorAnswer ?>

    I tried the above if statement which is checking for NULLS with == as well and I get the same thing. I also tried this: 

    
  if (!$results) {
   
 $errorAnswer = 'That Id was not found';
} else {
    $errorAnswer = '';
}

    Which doesn't work. Doesn't echo anything

    Many thanks

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted

    I did that, changed it to: 

    while(list($q, $c, $i) = $results->fetch(PDO::FETCH_NUM))

    after adding the SQL In:  

    SELECT q.question, o.choice, q.image 
    FROM questions q 
    JOIN quiz_options o 
      ON q.qid = o.qid 
    WHERE q.qid = 1;

    And print_r($results); gives me just the SQL: 

    PDOStatement Object ( [queryString] => SELECT q.question, o.choice, q.image FROM questions q JOIN quiz_options o ON q.qid = o.qid WHERE q.qid = 1;

    I was expecting array data.

    var_dump($i); gives me: 

    NULL

    and echo $i; does not print anything

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted 

      <?php

include 'includes/db.php';
$sql = "SELECT q.question, o.choice 
    FROM questions q 
    JOIN quiz_options o 
      ON q.qid = o.qid 
    WHERE q.qid = 1;";
$results = $pdo->query($sql);
// this produces rows that contain a question and a choice for that question.
// That means you have to handle the multiple choices by NOT showing the same question over and over.

$last_q = 'x';    // a non-existent value
$choice_num = 0;
while(list($q, $c) = $results->fetch(PDO::FETCH_NUM))
    // retrieve the contents of a single row as 2 vars
{
  if($q <> $last_q) // starting a new question so close the last div and begin a new one
  {
    if ($last_q <> 'x') // is this the first question? If not, close the last question's box
      echo "</div>";
    echo "<div class='question_box'>";  // start the new question box
    $choice_num = 0;  // new question; new choice count
  }
  $choice_class = 'choice' . $choice_num; // setup the appropriate choice class
  $last_q = $q;   // remember the current question
  echo "<p>$q</p>"; // show the question  - figure out later how to prevent repetition (which I did earlier)
  echo "<div class='$choice_class'>";   // a box for this choice
  echo $c;
  echo "</div>";    // close this choice's box
  $choice_num++;    // bump the choice cnt
}
echo "</div>";    // close the last question box

?>

     

    4 minutes ago, ginerjm said:

    accidentally or on purpose.

    lol, why would I do it on purpose?

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted

    18 hours ago, ginerjm said:

    If you don't want to pursue this effort I understand.  It's your task and your option and I'll drop out.

    No I appreciate the code and I will implement it to see what it's like but I just remembered why I wanted 4 separate elements for the choices. I want to model the quiz to be similar to this one:

    https://www.bbc.co.uk/news/world-64825206

    You see there is 3 choices, but mine will be 4 and you get the DOM scripting effect when You choose the correct answer and it goes green. Then you get green tick with a "Correct!". But I don't want the extra text underneath  to explain why the answer was right (or wrong) 

    But if you choose the wrong answer, you get a red X with "Wrong!" and then the correct answer goes green. But I don't want the percentages of how many people chose the choice.

    Should have mentioned all of this earlier.

     

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted

    This was the original idea: 

        <div class="quizContent">
  <p>Quiz goes here</p>
  <div class="answerA">
    <p>Answer A</p>
</div>
<div class="answerB">
    <p>Answer B</p>
  </div>
  <div class="answerC">
    <p>Answer C</p>
  </div>
  <div class="answerD">
    <p>Answer D</p>
  </div>
</div>

     

    3 minutes ago, kicken said:

    Do you actually need the separate classes though

    hmm....never really thought of it that way...just have all the answers in one div class and use css to style them as if they were in separate containers.

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted

    15 minutes ago, ginerjm said:

    If your foreach is looping thru the query results what is that "key=>value" pairing supposed to represent? 

    Which loop are you referring too? I was following mac_gyvers advice in he's earlier post.

    16 minutes ago, ginerjm said:

    And - Why do you need a unique class definition for the different question answers?  Can't you just use one class?

     

    For layout/CSS purposes 

    <?php

include 'includes/db.php';



$sql = "SELECT questions.question, quiz_options.choice FROM questions JOIN quiz_options ON questions.qid = quiz_options.qid WHERE questions.qid = 1;";

$statement = $pdo->query($sql);
$questions = $statement->fetchAll(PDO::FETCH_GROUP);



?>

<?php
foreach($questions as $question=>$group)
{
  ?>
  <div class="quizContent">
  <p><?php echo $question;?></p>

  <?php 

  $choice_styles = ['answerA', 'answerB', 'answerC', 'answerD'];
 foreach($group as $index=>$row)

  { ?>

         <?php
           
            echo "<div class='$choice_styles[0]'>  ".$row["choice"]." </div>";

         
         ?>

    
    <?php 
  
  }
 
  ?>
<?php 
 }
 ?>

    The rest of it is just HTML

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted

    Hi Guys,

    I've got most of the code done, I just can't figure out how to increment the array to output div classes: answerA, answerB etc

    This is my code: 

    <?php
foreach($questions as $question=>$group)
{
  ?>
  <div class="quizContent">
  <p><?php echo $question;?></p>

  <?php 

  $choice_styles = ['answerA', 'answerB', 'answerC', 'answerD'];
 foreach($group as $index=>$row)

  { ?>

         <?php
           
            echo "<div class='$choice_styles[0]'>  ".$row["choice"]." </div>";

         ?>

    
    <?php 
  
  }
 
  ?>
<?php 
 }
 ?>

    I've tried using counters, putting a ++ right there in the array index position but can't get it to work. Just outputs 4 x answerA classes.

    Many thanks

    How to show multiple records with the same name on the same page?

    in PHP Coding Help

    Posted

    14 hours ago, ginerjm said:

    Curious why you think you need a loop to add the css/html.

    Yes, your right. I only realised that after logging off. I initially tried to echo the HTML inside your while loop (just in case I could use the same loop) , got about 30% of the way there then logged off. 

    Thanks mac_gyver. 

    I should have pointed out I am only displaying one question at a time, but I just added a WHERE clause at the end of the SQL.

    12 hours ago, mac_gyver said:

    to apply the answerA, answerB, ... classes to each choice section

    mac_gyver, I have implemented your code and am trying the above advice now.  

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