Select from Mysql Database until a specific row problem

Question

I have two MySql tables for all of my Questionnaries as following:

Table 1:

id, quiz_name, quiz_no.

Table 2:

id, quiz_no, question, ans1, ans2, ans3, ans4, ans5.

Regarding the tables, I want to output data in separated divs with name of quiz on top and all the question and answers rows which belong of specific quiz_no (meaning to the specific quiz title) below. But all I got is all the rows with all the questions and answers below the same quiz_name at every title.

My code is:

quiz.php

$table1 = "SELECT * FROM table1";

if (!empty($_POST['num'])) {
  $no = $_POST['num'];
  $table2 = "SELECT * FROM table2 WHERE quiz_no = '".$no."'"; //Notice: Undefined variable: table2 in C:\... quiz.html
}
// So
$table2 = "SELECT * FROM table2";

quiz.html

<?php
      if($mysqli->multi_query($table1)){do{$result = $mysqli->store_result();
        while($row = $result->fetch_assoc()){
?>
        <div id="head_view_pers" class="modal-header" style="cursor:pointer;background-color:#7F4532;border:1px solid #7F4532;">
            <h4 style="font-weight:bold;color:#fff;margin-left:10px;" class="modal-title"><?php echo $row2['nume_chestionar'];?></h4>
            <i class="fas fa-expand" style="position:relative;top:6px;float:right;padding:0 10px 10px 10px;font-size:30px;font-weight:bold;color:#fff;"></i>
        </div>
        <article id="view_quiz_pers" class="view_quiz_pers" style="">
             <div style="padding: 0;" class="view_chestionare_pers">
                  <div class="div_title" style="border-bottom: 1px solid #000000;margin:10px 24px 15px 24px;padding-bottom:4px;font-size:28px;text-align:center;font-weight:bold;"><?php echo $row['quiz_name']; ?></div>

                  <form action="quiz.php" method="post">
                       <input type="hidden" id="num" class="num" value="" name="num[]"></input>
                       <div class="view_pers_wrapper" style="clear:both;text-align:justify;margin:0 40px;">
<?php                    
                         if($mysqli->multi_query($table2)){do{$result2 = $mysqli->store_result();
                              while($row2 = $result2->fetch_assoc()) {
?>
                                <div class="view_questions" style="clear:both;padding:0 0 10px 0;font-weight:bold;list-style-type:none;margin:0 24px;"><?php echo $row2['id']; ?>. <?php echo $row2['question'];?></div>
                                  <ul class="view_answers" style="clear:both;margin: 0 24px;list-style-type:none;">
                                    <li class="li_ans1" ><?php if($row['ans1']) {echo '<b>a. </b>'.$row['ans1'];} ?></li>
                                    <li class="li_ans2" ><?php if($row['ans2']) {echo '<b>b. </b>'.$row['ans2'];} ?></li>
                                    <li class="li_ans3" ><?php if($row['ans3']) {echo '<b>c. </b>'.$row['ans3'];} ?></li>
                                    <li class="li_ans4" ><?php if($row['ans4']) {echo '<b>d. </b>'.$row['ans4'];} ?></li>
                                    <li class="li_ans5" style="padding-bottom:24px;"><?php if($row['ans5']) {echo '<b>e. </b>'.$row['ans5'];} ?></li>
                                  </ul>
<?php 
                        }mysqli_free_result($result);}while($mysqli->more_results() && $mysqli->next_result());} 
?>
                                </div>
                        </div>
                  </form>
        </article>
<?php 
      }mysqli_free_result($result2);}while($mysqli->more_results() && $mysqli->next_result());} 
?>

EDIT:

<?php
 include 'db.php';

 $table1 = "SELECT * FROM table1";

 if (!empty($_POST['num'])) {
   $no = $_POST['num'];
   $table2 = "SELECT * FROM table2 WHERE quiz_no = '".$no."'"; //Notice: Undefined variable: table2 in C:\... quiz.html
 }

*The quizzes are inserted in the desired order. In the insertion page you can add only one questionnaire at time.

Here is the MySql database with the two tables: https://ibb.co/k624S82

Are the questions in the desired order in the database table? If not, then you might want to add a field such as question_no to specify the order and then order the results using that column. — Omari Celestine
– Omari Celestine, Commented Mar 3, 2019 at 12:42
Also, is the id of the questions an auto incrementing integer? — Omari Celestine
– Omari Celestine, Commented Mar 3, 2019 at 13:03
the questions are on the same row with the answers and each row has auto increment id — Gabriel
– Gabriel, Commented Mar 3, 2019 at 13:12

DinoCoderSaurus · Accepted Answer · 2019-03-03 17:18:09Z

1

I think the clue is in the code comments. This query
$table2 = "SELECT * FROM table2 WHERE quiz_no = '".$no."'"; seems to be what you want, because it "joins" table2 with table1. Unfortunately as you have noted, it gives this error
//Notice: Undefined variable: table2 in C:\... quiz.html.

It got replaced with this $table2 = "SELECT * FROM table2"; to mitigate the error. Which it did, ie it no longer gives the error, but it causes the next problem: not giving the correct result. It is a scope problem: since $table2 is declared inside the if block, it is only available in that block. Moving the table2 select outside the if was a good start, but now it doesn't "join" the tables.

This should give you the information you need to make progress on the problem.

answered Mar 3, 2019 at 17:18

DinoCoderSaurus

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Comments

imposterSyndrome · Accepted Answer · 2019-03-03 17:46:25Z

0

You are sending the value in an array in your form name="num[]" as such I suspect that your query doesn't work, and your undefined variable come from the query not being set at all,

$no = $_POST['num'];
  $table2 = "SELECT * FROM table2 WHERE quiz_no = '".$no."'";<-- you are trying to concatenate an array as a string

and it's then overwritten anyway as mentioned by @DinoCodeSaurus

What you need to do is get the value out of your post array ($_POST[num][0]) or, what you probably mean to do is just set it to a single value anyway and chage !empty($_POST['num'] to isset($_POST['num'])

unless i'm missing something

answered Mar 3, 2019 at 17:46

imposterSyndrome

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